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February 11

Depth of soil

Why in somewhere like England, is the depth of soil so great even on the sides or tops of hills? 92.29.123.158 (talk) 00:48, 11 February 2011 (UTC)

We have an article on the Geology of England which might be a good place to start. Vespine (talk) 01:13, 11 February 2011 (UTC)

... and, of course, the soil is extremely thin or even non-existent on the tops of some hills, especially in the Lake District, but peat explains the depth on many hilltops. Dbfirs 07:51, 11 February 2011 (UTC)

As the geology article mentions, it's partly to do with glaciers, but also why would you expect there not to be soil there? If plants can grow somewhere, which is the case in all of the UK thanks to its relatively warm and damp climate, then eventually soil will begin to build up, even on bare rock. Primary succession and ecological succession aren't particuarly good articles, but do discuss this. SmartSE (talk) 10:28, 11 February 2011 (UTC)

When was astigmatism first detected and corrected by spectacle lenses

I have invented a new type of spectacle frame and patented it, my reason was to enable spectacle lenses to be fitted which include astigmatism for easy construction in developing countrys. Currently best spherical correction is being used by NGOs globaly but its not providing a 100% accurate solution and 10% of the worlds population cannot work because they are refractivly blind. I will be envolved with a TV program made by the BBC shortly and I need to establish who and when astigmatism was first corrected in spectacle lenses. Any help would be appreciated. John D Snelgrove FBDO 86.167.248.127 (talk) 02:42, 11 February 2011 (UTC)

This book seems to indicate that astigmatism in the eye was independently discovered by Thomas Young in 1800 and later by George Biddell Airy in 1825. Airy is the one who coined the term, as suggested by William Whewell (who also, incidentally, coined the term "scientist"). The first lens that corrected astigmatism was developed in 1824 at Airy's request. As far as pat histories go, this one seems plausible enough to me. Unfortunately you cannot view the citations on Google Books so it is hard to know what the author is basing it on. --Mr.98 (talk) 03:54, 11 February 2011 (UTC)

In the Bakerian lecture of Young he describes his experiments on his own astigmatism. --Stone (talk) 07:30, 11 February 2011 (UTC)

Articel on the history of astigmatism.--Stone (talk) 07:38, 11 February 2011 (UTC)

The lecture of Airy is also available. --Stone (talk) 08:27, 11 February 2011 (UTC)

The Sun during solar maximum

What does the Sun look like through a (properly filtered) 13-cm telescope during solar maximum? I became interested in astronomy in my early teens (and bought a 13-cm telescope), when the Sun wasn't terribly exciting to look at because it was near a solar minimum. I'm much older now, but the Sun is STILL in the same minimum, so I'm very curious about what it would look like when it finally becomes active. --140.180.1.16 (talk) 08:40, 11 February 2011 (UTC)

First, let me slap on the usual DANGER! DO NOT OBSERVE THE SUN THROUGH A TELESCOPE! warning. Now, if you are "doing it right," you should never even look through a filtered telescope either. You should set your telescope up as a heliostat and project the image of the sun onto an imaging plane - do not use your eye as the focal plane! This is significantly less risky than just using a filter, and produces better images.

Usually, during solar maximum, the only thing you'll notice is "more sunspots;" but if you're talented at astrophotography, and heavily invest in specific filters, you can observe the chromosphere and the corona. If you're lucky, you may even catch a coronal mass ejection. CMEs can occur at any time, but are much more likely during solar max. To photograph a CME will be difficult. Historically, scientists waited for a total solar eclipse before attempting to photograph the corona; but modern optical equipment and film emulsions can make it possible to attempt chromosphere and corona photography directly, using a completely opaque sunshade. Here are some photos from NASA: Coronal Mass Ejections... (the photos came from the High Altitude Observatory at Mauna Loa Solar Observatory. Never fear; not everyone has access to such equipment. Here's a good "tutorial" that used a 4-inch refractor and a fairly "amateur" 35mm film camera: Observing and Photographing the Solar Chromosphere Through a Solar Cycle. That paper is a little bit old, so they describe using film (instead of digital) cameras; even with filtering, I'd be very nervous connecting my digital camera up to a heliostat, for fear of permanently burning/damaging my sensor. If you know your optics are safely filtered, and have a way to meter the light intensity, you could use a digital camera too. (Do not use your eyes to test your telescope's filter! Permanent eye injury/damage is a serious risk). Nimur (talk) 11:36, 11 February 2011 (UTC)

A solar filter, which often costs about $50+ USD (full-aperture, NOT eyepiece type, these can crack!!) is usually adequate for viewing the Sun safely by direct optical means, though a small hole or tear could completely compromise safety. A 13-cm telescope is large enough for the Sun's focused rays to do serious damage to the primary and secondary mirrors (assuming reflecting telescope) when using the projection method, and pointing it at the sun is dangerous as well for the finderscope if left uncovered. It is much safer to use both a solar filter and a brightness-reducing eyepiece filter in combination to view the Sun or otherwise leave the outer dust cover on if your telescope has both an outer ring and inner circle objective cover and then apply either projection or aperture-filtered observation, but test it using your camera first to check for any overexposure, although a specially-designed solar telescope is often specifically suited for solar observation but can be far more expensive than the combination of a usual telescope and a solar filter. ~AH1^(TCU) 22:08, 11 February 2011 (UTC)

Thanks for the info. One correction: the projection method definitely doesn't produce better images. There's a reason that people buy H-alpha filters even though they cost thousands of dollars. Even with a simple attenuation filter, you get much higher contrast than by projecting onto a piece of paper. --140.180.0.75 (talk) 01:48, 12 February 2011 (UTC)

By "better", I meant "bigger image." (Here's an example setup with a 3 foot projection from Sky & Telescope). The sun is huge and bright - so it sustains ridiculous magnification factors; and even with a small telescope, you should have enough resolution to capture a lot of detail when you project up to a 1-foot wide solar image. Now, you'd lose that resolution if your camera sensor were directly in the focal plane unless you've got a full-frame imager and tightly packed pixels and you have the optics to precisely control the image projection on your sensor. Depending on the effective field of view of your camera sensor, and the size of your primary, you probably aren't capitalizing on all available optical resolution of your telescope. "Best" results will of course depend on your intent and your available equipment. Nimur (talk) 18:15, 14 February 2011 (UTC)

Effect of damaging the Earth's core

Not long ago, some classmates presented a plan to dig a tunnel through the Earth's core to provide a high speed train route between New York City and Tokyo. They were unable to answer the question of whether this tunnel would affect the planet's structural stability. Would a tunnel through the core increase the damage that asteroid impacts, etc., could cause? Also, if terrorists managed to detonate a suitcase nuke within the core, what size of earthquake could they cause? (I ask because in 2009, South Korea reported that a North Korean underground nuclear test had created a magnitude 9 earthquake.) NeonMerlin 10:34, 11 February 2011 (UTC)

There is no conceivable way that a train-sized tunnel would affect the "structural stability" of Earth. What you would need to worry about is whether the tunnel could be structurally sound. But analyzing the hypothetical tunnel's stability is moot - there is no current technology that could be used to dig such a tunnel. Even if the tunnel were not "direct", but instead followed the contour of the earth's surface, we do not have a way to safely excavate a "subway" tunnel underneath the depths of the open ocean across the Pacific. Finally, even if a nuclear weapon could be detonated at great depth (say, even 100 km below the surface, using a hypothetical tunnel or any other proposed mechanism) it is unlikely that it would trigger a catastrophic earthquake. Scientists have been measuring the seismic behavior of nuclear weapons since 1945; the media often grossly misrepresents the energy scales involved. A group at CISAC produced this public report, Technical Analysis of the DPRK Nuclear Test. You may want to re-check where you heard your "magnitude 9" claim; I'm pretty sure that is an inflated number; but it's really irrelevant anyway, because there are so many technical details involved in converting measurements of a bomb into an "equivalent-sized earthquake." The energy released by even the most powerful nuclear bombs is dwarfed in magnitude by even a small earthquake. Nimur (talk) 11:51, 11 February 2011 (UTC)

(EC) The structural damage is probably negligible - a tunnel is tiny compared to the size of the earth (or the core). For the same reason, a nuclear explosion in the core would not or hardly be noticeable up here. Aside from the engineering difficulties of protecting the tunnel from the stresses and temperatures prevalent in the core, are you aware of the kind of slope the train has to go down and up again when it is near the surface? I'm too lazy to calculate it but it'll probably be more than 45 degrees, much more than any normal train can handle. --Wrongfilter (talk) 11:54, 11 February 2011 (UTC)

49 degrees, according to my calculations assuming a spherical cow Earth. A straight-line tunnel would reach a depth of about 2200 km, penetrating far into the lower mantle, but missing the core by some 500 km. –Henning Makholm (talk) 13:06, 11 February 2011 (UTC)

And, as we all know (I assume ;-), a spherical train in a frictionless vacuum moving under the influence of gravity only would need 49 (IIRC) minutes for the trip, regardless of the distance covered. --Stephan Schulz (talk) 14:30, 11 February 2011 (UTC)

Schults, that's only true if the shape of the tunnel is a cycloid and the density of the earth is uniform, which it isn't.Dauto (talk) 16:01, 11 February 2011 (UTC)

Uniform density and non-rotating Earth are, of course, part of the spherical cow universe! --Stephan Schulz (talk) 19:55, 11 February 2011 (UTC)

There's one, erm, small technical obstacle to this plan. The inner core rotates relative to the Earth's surface, something like once every 400 years if I remember correctly. As for nukes, they can be detected on a seismograph, but they are very small. Unfortunately Comprehensive Nuclear-Test-Ban Treaty Organization doesn't get into the specifics of the amplitude, but it's something like 3. There's a formula for the size of the bubbles of vaporized rock from a nuke at Underground nuclear testing - but it's something like hundreds of feet near the surface, and in the core I bet the pressure would reduce the size further. Wnt (talk) 18:07, 11 February 2011 (UTC)

Someone mentioned above that this hypothetical tunnel is supposed to miss the core by some 500 kilometers so the rotation of the inner core wouldn't be a problem, if that happens to be correct. Dauto (talk) 00:38, 12 February 2011 (UTC)

Something else to mention is that plate tectonics all occurs in the Earth's crust, so an explosion far below the crust is unlikely to have much effect. I would suspect that the right-sized nuke, placed at the hypocenter (focus) of the potential earthquake, a few miles underground, typically, could trigger a quake. However, this would only work if strain had built to the point where a quake would occur soon, anyway, and the quake would be no bigger than the normal range for that fault (it might actually be smaller, if it occurs somewhat earlier, with less strain accumulated). Consider an analogy with cloud-seeding to make rain. It doesn't work when there's no moisture in the air, it has to be almost ready to rain anyway. StuRat (talk) 01:20, 12 February 2011 (UTC)

Actually, the base of the lithosphere (and all plates) is in the uppermost mantle, the asthenosphere, but you're right that you can't cause an earthquake in that way, only trigger it if it's already near a critical point in the earthquake cycle. There are often foreshocks before an earthquake, but opinion is divided as to whether they actually trigger the main event or are just a manifestation of the increasing stress levels in the area where the mainshock happens. Current thinking is leaning towards the latter explanation suggesting that it's actually difficult to trigger an earthquake before it's ready to go. Mikenorton (talk) 10:13, 12 February 2011 (UTC)

Castrating cats

Is it right to castrate domestic cats to make them stay at home? — Preceding unsigned comment added by Sina-chemo (talk • contribs) 11:57, 11 February 2011 (UTC)

The reference desk doesn't give moral judgements; it only answers factual questions. The article Neutering has information on the positive and negative effects of neutering a cat (but please consult a vet if you have specific questions as we can't give veterinary advice). --Colapeninsula (talk) 12:31, 11 February 2011 (UTC)

I don't know if people do it to make it stay at home. I always thought it was to avoid it getting nuts. Wikiweek (talk) 12:42, 11 February 2011 (UTC)

It helps keep the cat from getting testy. Googlemeister (talk) 14:19, 11 February 2011 (UTC)

No kitten; it really works. Matt Deres (talk) 14:52, 11 February 2011 (UTC)

Purrr-fect answers. 10draftsdeep (talk) 16:05, 11 February 2011 (UTC)

I get the feline you're treating this like some kind of joke... Vimescarrot (talk) 18:01, 11 February 2011 (UTC)

Please, don't make catty comments on the ref desk. It pusses me off. Tinfoilcat (talk) 19:07, 11 February 2011 (UTC)

Surely a better reason to castrate cats is to avoid getting too many more cats. HiLo48 (talk) 21:46, 11 February 2011 (UTC)

See feral cat. ~AH1^(TCU) 22:01, 11 February 2011 (UTC)

Uncastrated toms are rather objectionable, doing things like spraying their "scent" on the drapes. If I had such a cat, the only choice would be between castration and putting it to sleep. If it could make the choice, I think it would prefer to skip the nuts. StuRat (talk) 01:03, 12 February 2011 (UTC)

It is a good idea to castrate them. It doesn't make them stay at home, but it makes them roam, fight and mate less. And, of course, they will not produce unwanted offspring. You should do this to your cats unless you are breeding them. Zzubnik (talk) 14:57, 14 February 2011 (UTC)

Castration would seem an in effective method of birth control. Unless every single tom in a neighborhood is castrated, a queen in heat will find a willing an capable mate. Spay (ovario-hysterectomy) 50% of the stray queens and you cut the birth rate in half. Castrate 95% of the toms and you have no effect at all. -- 119.31.126.67 (talk) 00:34, 16 February 2011 (UTC)

differentiation by first principles

how to obtain derivative of e to the power root x by using limits? —Preceding unsigned comment added by 180.215.34.14 (talk) 12:28, 11 February 2011 (UTC)

This is a duplicate of a recent mathdesk question, probably by the same anonymous user: WP:RD/MA#differentiation by first principles. Please don't ask the same question on several desks. –Henning Makholm (talk) 12:44, 11 February 2011 (UTC)

If you read closely, you'll see that this is actually a different question, in that it adds the specification "...by using limits". The answer given on the math ref desk used various rules about differentials, rather than using just the definition of the derivative as a limit, and expressing a derivation of the result by using expressions involving limits. Red Act (talk) 22:40, 11 February 2011 (UTC)

 Please solve USING LIMITS.  —Preceding unsigned comment added by 180.215.44.196 (talk) 07:55, 12 February 2011 (UTC) 

It follows by definition if you choose characterization 4 from Characterizations of the exponential function! Anyway the maths reference desk is the right place. Dmcq (talk) 12:13, 12 February 2011 (UTC)

Extra 4 minutes

If a day is 23 hours and 56 minutes long instead of exactly 24 hours, how do the extra 4 minutes fit in? jc iindyysgvxc (my contributions) 12:52, 11 February 2011 (UTC)

It's the difference between the solar day and the sidereal day. The Earth rotates once every 23 hours and 56 minutes with respect to the stars. But it also circles the sun (or, geometrically equivalently, the sun circles the Earth) once per year. So the Earth has to do 4 minutes worth of catching up to the relative movement of the sun every day. Note that 365*4 minutes is nearly exactly one day, i.e. the extra rotation due to the motion around the sun. --Stephan Schulz (talk) 13:00, 11 February 2011 (UTC)

In other words, if we measure the day by tracking the location of the sun, we would have to wait an extra 4 minutes before we turn a full 360 degrees. And that is the source of the discrepancy. That's because the earth has moved along a small distance in its orbit of the sun between today and yesterday. --Jayron32 13:46, 11 February 2011 (UTC)

Thanks. I've heard the term "sidereal" and read of the difference but never had it explained so clearly in terms of "one day per year." Edison (talk)

Supposing for the sake of simplicity that the year was exactly 365 days long, the sidereal day would be exactly 365/366 of the solar day, and the number quoted above as "4 minutes" would turn out to be be exactly 1/366 of a year. --Anonymous, 03:40 UTC, February 12, 2011.

Leap years

then why do we leap 1 day every 4 years? —Preceding unsigned comment added by 165.212.189.187 (talk) 16:52, 11 February 2011 (UTC)

That's another matter entirely. We leap a day every so often because a year is actually slightly longer than 365 days. Dauto (talk) 17:05, 11 February 2011 (UTC)

In other words, it takes about 365 1/4 turns on the earth's axis to arive at the same point in the earth's orbit. That extra 1/4th of a day adds up and needs to get tacked on every 4 years. --Jayron32 18:09, 11 February 2011 (UTC)

Specifically, 97 times every 400 years because it takes 365.24 turns.  :) —Preceding unsigned comment added by 205.193.96.10 (talk) 20:45, 11 February 2011 (UTC)

The "one day per year" is because 364 days are due to the Earth turning, and the last one is because we are going around the sun. If the Earth didn't rotate at all, we would still get one day per year due to us going around the sun. Leap days, on the other hand are caused because it takes an extra 0.24 days to get around the sun and start the seasons over; it has nothing to to with the Earth's own rotation. —Preceding unsigned comment added by 205.193.96.10 (talk) 20:49, 11 February 2011 (UTC)

Actually we have 366 (plus about a quarter) days due to the earth's rotation and (-1) days due to the earth's motion around the sun for a total of 365 (plus about a quarter). Dauto (talk) 00:11, 12 February 2011 (UTC)

See leap year and leap second. The sideral motion of the Moon also varies in this matter as it catches up to the Earth on one side of its orbit while remaining gravitationally locked to face one side to Earth. ~AH1^(TCU) 21:59, 11 February 2011 (UTC)

Radio power

Hi, does anyone know how much radio signal power is needed in order to pick up a station on an ordinary household radio (with no special aerial or other fancy equipment)? I'm talking about the power "in the ether" at the place where the radio is located, not the total power of the transmitter, which is at an unspecified distance. I think I'm looking for an answer in watts/m^2 (if not, please correct me). Does the answer differ much between FM and AM? 86.179.4.118 (talk) 14:39, 11 February 2011 (UTC)

The integrated power at the aerial terminal is given (albeit without source) in Orders of magnitude (power) as on the order of femtowatts for an FM signal and a good-quality radio receiver, but I'm not sure what the appropriate 'effective area' would be for the aerial. A similar figure (roughly 10 fW) is given for a minimum receivable spread-spectrum cellular telephone signal. As a very rough (order of magnitude) estimate, a detectable FM signal flux is going to be on the order of femtowatts to picowatts per square meter, depending on the size and configuration of your antenna. TenOfAllTrades(talk) 15:24, 11 February 2011 (UTC)

See "Field strength." The "signal power" of a broadcast signal at the receiver location is typically given in units of microvolts per meter, rather than in watts per square meter. Then the signal at the radio's antenna terminal or first RF input stage is affected by the type and orientation of the antenna, or the "gain" of the antenna. Modern radios often use a little ferrite loopstick in the radio, while older ones used a flat coil on the back panel, and older ones still, with less RF amplification, required a long overhead antenna and an Earth ground. Naturally a weaker signal or stronger signal will produce an audio output with varying signal to noise ratio, and with varying clarity compared to atmospheric static, interference from electrical equipment, thermal noise in the receiver circuitry and interference from other stations on the same or adjacent frequencies. The bandwidth affects the signal to noise ratio. A signal to noise ratio of 3 deciBels is sometimes described as the minumum usable signal, but it would sound horrible. A 10 dB SNR is often used in describing the sensitivity of a receiver, per [1]. Someone with recent skills or training in communications calculations might be able to come up with the microvolts per meter for a 10 dB Signal plus noise to noise ratio on a typical (say AM) receiver in the broadcast band. I once saw a map of Italy showing the signal strength at varying locations in that country, regardless of what the transmitter was, but I haven't seen that for the US. Says that the sensitivity of modern radio receivers ranges from several microvolts to several millivolts. I suppose that to convert this to "power" you would need to know the input impedance of the radio at the antenna connection, which for most ordinary AM radios is somewhere internally where the loopstick connects to the first RF stage. Edison (talk) 16:00, 11 February 2011 (UTC)

You can approximately convert electric field strength (volts/meter) to radiant intensity (watts per square meter) by squaring and dividing by the antenna impedance; in practice, you need to account for the antenna matching properties, impedance of free air, antenna directivity, and so on. A comprehensive theoretical treatment and several useful conversion tables are provided by Antarctic Impulse Transient Antenna website at University of Hawaii: Field Intensity and Power Density, which is an excerpt from the Navy Electronic Warfare & RADAR Systems Handbook. Nimur (talk) 22:14, 11 February 2011 (UTC)

James Watt steam engine patent

I am looking for the patent that James Watt filed in 1784. This patent is mentioned in [article "History of Rail Transport"], but I cannot locate an electronic copy of the patent itself. Thanks. 128.223.222.16 (talk) 16:30, 11 February 2011 (UTC)

I don't see it cited there or at James Watt#Early experiments with steam or even anything specifically about the 1780s at all in Watt steam engine. That's bad. DMacks (talk) 19:25, 11 February 2011 (UTC)

I looked briefly at some old Google Books entries about Watt's 1784 patent; it seems to be a real thing, but none of them gave patent numbers. The European patent office search doesn't seem to let you search very old patents very easily, even though they may be on there (like the 1769 Watt patent linked in our article). Seems like it is not going to be a very easy thing to dig up online. --Mr.98 (talk) 21:54, 11 February 2011 (UTC)

Not sure whether Google Patents goes that far back, but this link does mention the invention and existence of a patent. ~AH1^(TCU) 21:57, 11 February 2011 (UTC)

Google Patents only contains US patents. The US didn't issue its first patent until 1790. I've assumed the 1784 patent must be a British one, in part because his 1769 one is. --Mr.98 (talk) 00:00, 12 February 2011 (UTC)

Your local library may be the only source at the moment. According to Birmingham City Council the European Patent Office is currently embarked on the "digitisation of UK Patents 1870 - 1920", although this espacenet publication contradicts that and suggests that there's not much hope for anything before 1890.

By the way the number of the 1874 Watt patent is GB1432. --Heron (talk) 11:48, 12 February 2011 (UTC) Ignore all that. I misread the dates. Sorry. --Heron (talk) 12:15, 12 February 2011 (UTC)

Got it in three ! Project Gutenberg has Kinematics of Mechanisms from the Time of Watt by Eugene S. Ferguson which has a very detailed history of Watt's production, competition, and patents, including a diagram and description of "Watt's mechanisms for guiding the upper end of the piston rod of a double-acting engine (British Patent 1432, April 28, 1784)". SamuelRiv (talk) 20:19, 12 February 2011 (UTC)

Bengal fire

Resolved

What exactly is Bengal fire (a.k.a. Bengal light), and how did it get its name? All I found on the web so far is this – and many indirect uses, such as flowers and tea carrying that name. — Sebastian 21:21, 11 February 2011 (UTC)

This Everything2 post explains that it is a firework, mostly potassium nitrate, adulterated with copper and/or barium to add a blue-green color. Is is so named because these fireworks/explosives were originally originally manufactured and supplied from Bengal to England. The Everything2 node isn't quite a reliable source; and though it cites a few references, they are also of dubious quality. But this journal article, Pyrotechnics in Fireworks (2004), seems to corroborate the general claims, though. It seems that other "brightly colored" things use the term "bengal fire" because it sounds neat. Nimur (talk) 21:41, 11 February 2011 (UTC)

Thank you, Nimur! — Sebastian 22:20, 11 February 2011 (UTC)

Actually, as I'm trying to incorporate that information, I noticed that we used to have a redirect from Bengal fire to Bengal fire stick. I restored the latter in my user space so I can ask here if there's anything to it; it is at least not an obviously fictitious page, as the deleters assumed. — Sebastian 22:45, 11 February 2011 (UTC)

I recall purchasing "Bengal matches" around November 5th many years ago. Were they an early alternative to sparklers? Dbfirs 00:20, 12 February 2011 (UTC)

February 12

Boa constrictor scientific naming...

According to the Boa constrictor article - "Though all boids are constrictors, only this species is properly referred to as "Boa constrictor"; an almost unique instance of an animal having the same common and scientific binomial name". So, which came first - the common name, or the scientific name? --Kurt Shaped Box (talk) 04:15, 12 February 2011 (UTC)

I take that sentence to mean that it has always been known in English as a boa constrictor. Since that is its common name and its scientific name how can one say that either came first? (Of course its South American name jibóia was presumably used before that.)--Shantavira|^{feed me} 07:57, 12 February 2011 (UTC)

Linnaeus created this name in 1788, and he wrote in Latin, so if we equate "scientific binomial name" and "Latin name given by Linnaeus" then it would appear that the scientific name came first. The Penny Cyclopedia in 1836 referred to the name as "popular", so the transition from scientific to popular came between those two dates. The Latin word boa, incidentally, goes all the way back to Pliny. [Source: OED] --Heron (talk) 11:23, 12 February 2011 (UTC)

There are other species known in English by their binomial name. Aloe vera is another one. Dauto (talk) 15:57, 12 February 2011 (UTC)

In some cases, it would be redundant, as in Gorilla gorilla! --FOo (talk) 18:13, 13 February 2011 (UTC)

when the earth was shining

the first earth surface might be very hot between 2000~3000 degrees centigrade . this temperature cases the rocks be melt then this hot cases the matter such as the volcanic magma to shine . before the cooling surface of earth and when the sun was young it is clear that the earth was red color and shining . how much years was the during of this condition . (all remarks are collected from astronomic articles)--78.38.28.3 (talk) 04:54, 12 February 2011 (UTC)a. mohammad zade Iran 2011

We don't know very much about the Hadean period, but our article History of the Earth might be of interest. If the Earth initially shone, then it could have done this for only a comparatively short time, perhaps between 4,540,000,000 and 4,530,000,000 years ago (when the moon was formed and the Earth had a crust). Dbfirs 10:33, 12 February 2011 (UTC)

Use of the word "shine" in this context is potentially confusing, since it is frequently used in astronomy, as elsewhere, to denote the reflection of incident light, as in Earthshine. Colloquially, all the planets (and their satellites), including the Earth, "shine" by reflecting the light of the Sun (Earthshine being a double reflection), according to their individual albedos. For the phenomenon you are interested in, incandescence, or colloquially "glow", would be more appropriate. 87.81.230.195 (talk) 13:14, 12 February 2011 (UTC)

Which incident light does the Sun reflect to create sunshine, then? –Henning Makholm (talk) 15:22, 12 February 2011 (UTC)

Yes, that's all moonshine. Dbfirs 22:03, 12 February 2011 (UTC)

I said "potentially confusing", "frequently" and "more appropriate", not "wrong", "invariably" or "exclusively," and "solely appropriate". I was also pointing the OP towards articles on subjects relevant to his query, and providing hopefully polite guidance on vocabulary usages in a language in which he is evidently not fluent. How helpful do you suppose your comments were? *Stalks off in high dudgeon.* 87.81.230.195 (talk) 22:50, 12 February 2011 (UTC)

Yes, your reply was perfectly acceptable and you provided useful links, and we weren't really complaining about it, but we perfectly understood the usage of the OP whose first language is not English. Apologies for any offence caused

. Dbfirs 08:37, 15 February 2011 (UTC)

only one of my friends understood my meaning . I know that the earth is reflecting the sun shine all over the life of that and all matters such as dark matter has shining . the last planet that nasa discovers is melt and hot and is shining

• File:509304main kepler rocky planet full
• File:508881main KOI-72-STILL-6k-1 full full

so it can shine without reflecting the other star light--78.38.28.3 (talk) 04:51, 13 February 2011 (UTC) a. mohammadzade]] —Preceding unsigned comment added by 78.38.28.3 (talk) 04:46, 13 February 2011 (UTC)

Please see the articles on the Hadean period, Giant impact hypothesis, Cool Early Earth, and Late Heavy Bombardment. Because rocks from asteroids date back to 4.6 billion years, and rocks from Earth date back only to 3.8 billion years, it was thought the surface was largely molten for 800 million years, but now some people think that it was solid for most of this time, until a wave of asteroid bombardment. But according to Giant impact hypothesis, the existence of a magma ocean has never actually been proved at all! Unfortunately, the answer here is that it just isn't known yet. Wnt (talk) 07:55, 13 February 2011 (UTC)

Borosilicate glass

r light bulbs Borosilicate glass — Preceding unsigned comment added by Tomjohnson357 (talk • contribs) 05:15, 12 February 2011 (UTC)

If you're talking about ordinary incandescent lamps, then the outer envelope can be soft soda glass, which is cheaper to make and form. However, the inner glass stem that the metal supports are fixed to, are of borosilicate due to its low coefficient of expansion. Other lamps that run very hot, like vapour discharge lamps, are mostly borosilicate glass throughout.--Aspro (talk) 10:55, 12 February 2011 (UTC)

[2] this book says the same. The halogen lamps are made from high silica glass or aluminosilicate “hard” glass.--Stone (talk) 10:59, 12 February 2011 (UTC)

capacitance

4 μF capacitor is charged by 250V.Find energy stored in it.It is connected in parallel combination to an uncharged 3μF capacitor. Find energy stored. —Preceding unsigned comment added by 180.215.44.196 (talk) 07:41, 12 February 2011 (UTC)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our policy here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know.

Have you read our Farad article? It gives various equations that relate the charge, voltage, and stored energy to the capacitance. For the first question, use the appropriate equation directly. For the second question, the first capacitor will discharge (dropping its voltage) into the second until the voltage on both is the same; no charge is lost. Find the initial charge on the first capacitor, then find the corresponding volatage on the combined system. CS Miller (talk) 10:20, 12 February 2011 (UTC)

That sounds an awful lot like my homework from last semester. I already had the pleasure of doing it, I don't want to rob you of the same experience. Dismas|^(talk) 10:22, 12 February 2011 (UTC)

Radio activity

A radio active material at a given instant emit 3000 particles per minute,10 minute later it emits 1500 particles per minute. find decay constant and half life. —Preceding unsigned comment added by 180.215.44.196 (talk) 07:45, 12 February 2011 (UTC)

We avoid homework questions as if they were radioactive. Clarityfiend (talk) 07:55, 12 February 2011 (UTC)

I suggest you read decay constant and half life.--Shantavira|^{feed me} 08:36, 12 February 2011 (UTC)

Legal chemistry

Is there and article regarding legal chemistry?--Email4mobile (talk) 10:35, 12 February 2011 (UTC)

Probably forensic chemistry. Someguy1221 (talk) 10:39, 12 February 2011 (UTC)

Or forensic toxicology, as the original use of the term 'legal chemistry' was all about poisoning. Mikenorton (talk) 10:43, 12 February 2011 (UTC)

Thanks a lot. I will add Arabic interwiki to that articles.--Email4mobile (talk) 11:01, 12 February 2011 (UTC)

Changing the length of a second/minute/hour/day/year

The recent leap day related questions left me with a question that I don't think has been addressed. Has the idea of changing the official length of a day (or whatever would be necessary) so that we don't need as many/any leap days been seriously proposed? Or are we at an optimum now? Dismas|^(talk) 10:44, 12 February 2011 (UTC)

Well, if you change the length of the day, so that the year has exactly 365 days, then noon will be shifted from 12:00 to almost 18:00 within a year - I don't think that's acceptable. Icek (talk) 11:11, 12 February 2011 (UTC)

Exactly. The fundamental problem is that the length of the tropical year – essentially, the length of one full cycle of seasons – just isn't an integer multiple of the solar day. If we insist that our timekeeping system stays in sync with the sun (12:00 noon is always in the middle of the daytime) and with the seasons (the winter solstice will always fall in late December), then these little corrections are unavoidable. TenOfAllTrades(talk) 15:35, 12 February 2011 (UTC)

Proposal to abolish leap seconds have been put forward by people who don't understand the issue properly. I think the current arrangement will survive for sometime yet and is as good as is reasonable and workable. --Aspro (talk) 11:12, 12 February 2011 (UTC)

The second is an arbitrary human unit, unlike the solar day and tropical year, though in practice it's too entrenched to be redefined. As I software engineer I don't like the current system of random leap seconds announced six months in advance. It's akin to announcing leap years on March 1 of the previous year. I'd much prefer a predefined schedule. The schedule doesn't have to be "no leap seconds ever", it just has to be predictable. -- BenRG (talk) 18:18, 12 February 2011 (UTC)

That's why things are as good as can be expected . The change in the Earth's rotation can't be modelled very well yet based on our current understanding. Even changing the way a second is defined will not get around this fact. To expect more at present is just wishful thinking. --Aspro (talk) 18:47, 12 February 2011 (UTC)

The problem is that the need for leap seconds is inherently unpredictable; the rotation rate of the Earth just isn't regular enough for us to predict the need for a leap second well in advance. Between 1988 and 1998, there were eight leap seconds added; between 1998 and 2008, there was just one: File:Leapsecond.ut1-utc.svg. In principle, if one were willing to tolerate an increased drift in UTC versus mean solar time, one could implement a longer period between leap second adjustments or allow a larger deviation to accumulate before corrections were applied — but it still wouldn't prevent those adjustments from being at 'random' intervals. TenOfAllTrades(talk) 19:00, 12 February 2011 (UTC)

A perfectly workable solution would be simply to abolish leap seconds and let UTC float with respect to the Earth's rotation. The offsets between local mean solar time and civil time that we already accept as unproblematic everyday consequences of zonal time are far larger than the UTC drift that could accumulate until several centuries pass. And once those centuries do pass, countries can simply choose to switch to another UTC offset for their civil timekeeping. Many countries implement such switches twice per year already, and modify their rules for these changes relatively frequently. Simply skipping one of the diurnal switches in order to move one timezone east or west will be a fairly routine exercise, compared to the practical benefits a predictable "universal time" scale would bring.

Alternatively we could start defining civil time everywhere in terms of hourly offsets from TAI and abolish the entire concept of a separate UTC scale. This would be just as workable in the long term, but the half minute of TAI-UTC difference that has already accumulated would make the switchover impossibly chaotic as a practical matter, and for no real benefit compared to a floating UTC. –Henning Makholm (talk) 02:27, 13 February 2011 (UTC)

Floating universal time (GMT) was abandoned because it became impractical and got replaced by UTC. Satellite navigation may remove some of the current dependence on this, but one still has the problem of how to constantly synchronise all the independent time pieces at odd times throughout the year which may actually require more disruption than the present system. Transaction in commerce are completed and logged in milliseconds. Television transmission networks which are slightly out of sync falter. However, the suggestion to use ITA with offsets, would not only work, but is what is done now, but to let ITA loose and let it drift away as civil time and away from the astronomical time datum, would make future interpretation of data (especial astronomical data) more difficult. All the pros and cons added together leave us with what we have now, as the best compromise, and so we wont start going backwards on the basis of spurious arguments. --Aspro (talk) 12:38, 13 February 2011 (UTC)

Um, GMT does not float; on the contrary it is explicitly tied to the Eath's rotation by definition. In a sense, what is done now is to use ITA with offsets, except that those change unpredictably, leaving gaps and overlaps in the otherwise continuous time scale. Sure, astronomers need to keep accurate track of the Earth's rotation for their own particular reasons -- they shall be welcome to keep historical records about that so they can interpret their observations in the future -- but that is not a good reason to shackle all of the rest of humanity to an unpredictable, non-continuous, non-arithmetic civil timekeeping scale. –Henning Makholm (talk) 16:34, 13 February 2011 (UTC)

You say: Um, GMT does not float; on the contrary it is explicitly tied to the Eath's rotation. Please read Non sequitur (logic). See image . I don't mind explaining things but please don't come back with absolute tosh!--Aspro (talk) 18:16, 13 February 2011 (UTC)

A "floating" time standard is one that isn't tied to the Earth's rotation. -- BenRG (talk) 22:45, 13 February 2011 (UTC)

The question is, if you add $\lfloor {\tfrac {1}{2}}a(t-t_{o}+86400s)^{2}\rfloor -\lfloor {\tfrac {1}{2}}a(t-t_{o})^{2}\rfloor$ seconds to the end of every day, where t_o is circa 1820 and a ≈ 0.0014/day/century, what's the expected maximum drift from solar time over, say, the next thousand years? I don't know the answer, but if it's, say, ten minutes, then I think it would be a huge improvement over what we've got now. I'm pretty sure the number of people who don't need to correct for the current 1-second drift but would need to correct for a ten-minute drift is small compared to the number of people inconvenienced by the current system. For heaven's sake, we tolerate a 0.3% variation in calendar year length and a whopping 10% variation in calendar month length. Astronomers have to correct for those too. -- BenRG (talk) 22:45, 13 February 2011 (UTC)

The short term fluctuations are random and 1820 was the nadir of a parabola, which means the drift does not follow a liner increase - so that doesn't work. Recalculating every day or other regular period and adjusting, goes against every other method of standardization so that's not on either. --Aspro (talk) 19:55, 14 February 2011 (UTC)

I know it's a parabola; that's why there's a ² in the formula. I agree that it's unlikely to be adopted, though weirder things have happened. -- BenRG (talk) 21:27, 14 February 2011 (UTC)

Not to mention that we tolerate a ±30 minute deviation from solar time just so that our time zones can change on one-hour boundaries rather than one-minute or one-second. Why is one-second accuracy in UTC so important? -- BenRG (talk) 23:07, 13 February 2011 (UTC)

It is not about looking out the window a noticing anything different. The current time systems have evolved as technology has advanced to benefit from evermore sophisticated time standards. Leap seconds are one of those steps to bridge the gaps until the next advance is ripe for adoption. Abandoning them because some people don't care about the problems it will cause in multiple other places is not the way to go. For computer systems the Network Time Protocol adjusts for leap seconds automatically. Systems for which this approach is unsuitable, can stick with ATI as standard time reference.--Aspro (talk) 19:55, 14 February 2011 (UTC)

I'm aware of the existence of NTP and TAI. Other than that, I can't make any sense of your reply. Isn't "the next advance" what this crazy subthread is about? -- BenRG (talk) 21:27, 14 February 2011 (UTC)

Did this current system come about over-night? No, it didn't! From the time when the ancient Egyptians just counted the days and ignored the astronomical year, the advancement of the measurement of time has advanced stepwise. The introduction of leap seconds is just one of those many steps. At each step, some individuals have objected, as it has forced them to adopt new ways of doing things. I don't think there is any point in trying to explain this in simpler terms as you appear to be not trying to understand. A trait that others, who object to leap seconds, seem to share IMHO.And which you appear to be now confirming.--Aspro (talk) 21:54, 14 February 2011 (UTC)

±30 minutes is only the minimum necessary deviation. In practice, there are places that use civil timescales as much as 90 minutes from local mean solar, such as Galicia (Spain) (150 minutes in the summer), Iceland or Western Sudan. –Henning Makholm (talk) 00:34, 14 February 2011 (UTC)

Physics Entries in General

I am new to wikipedia. I am writing a book on evolution and my areas of expertise are biology, animal behavior and evolution. I am however writing an introductory chapter on atoms in order to define where life does not appear to exist ( in order to later identify the first molecules which could be considered to have "life"). As I have been reviewing the elementary particles such as quarks and gluons, etc I have found a very disturbing trend to not clearly define right up front which aspects of particles and theories have been clearly "proven" and observed and which concepts are theoretical in nature. I feel that even as not being a physics specialist, that this concept regarding scientific theories is extremely basic. I added some comments to the gluon page in this regard and they were removed. I have no wish to "soapbox" about this concept. But I have noticed for physics based articles, often the pages will immediately start going off on very mathematic based formulas without even stating whether there is any proof of the particle or concept. In biology I can talk about tree bark and vascular systems and photosynthesis and we can go in and very soundly prove and display and observe the science right down to the molecules which house the chemical reactions from sunlight to sugars. But the particle folks go into arcane mathematics which is almost religious in nature without ever clearly stating whether for example particle spin (1/2 or full integer) is theorized or has actually been observed. It is like smoke and mirrors with math being the mystical gate. In other words "regular" people cant consider the accuracy of these theories as they immediately made opaque by the insertion of mysterious terms and equations. I think this concept is very basic and extremely unscientific. How does one taker on these articles in order to demand very basic descriptions of what exactly is proven and what exactly is theoretical up front? Thanks so much for your time. — Preceding unsigned comment added by Maplelanefarm (talk • contribs) 14:47, 12 February 2011 (UTC)

Again..thanks for your time...I think this is a huge issue with most of the particle pages. — Preceding unsigned comment added by Maplelanefarm (talk • contribs) 14:49, 12 February 2011 (UTC)

Indeed, you do need to know mysterious terms and equations to make sense of the subatomic world. There isn't much of a way around that. There are experiments, say, that show that spin certainly exists. But even interpreting what is going on in the experiment requires you to buy into a lot of complicated maths. Most of the standard model is "proven" (as far as that goes) pretty definitively. The stuff that is wholly "theoretical" lies just on the edges — e.g., quantum gravity, string theory, etc. The presence of math does not indicate that something is not experimentally verified. As just an indication, the classic experiment proving the existence of spin is the Stern–Gerlach experiment. It doesn't rely on heavy maths, but it still requires quite a deep theoretical understanding to make sense of what is going on. I'm inclined to think this is an artifact of the physics, not so much Wikipedia, though I do agree that the high levels of technical jargon in such articles make them pretty impenetrable. --Mr.98 (talk) 15:15, 12 February 2011 (UTC)

Why do you say you can "prove...and observe the science right down to the molecules..."? The tools you use to do that such as electron microscopes and mass spectrometers all rely on physics concepts like particle spin and electron orbitals. To properly complete your "proof" you will need these concepts too. Franamax (talk) 15:56, 12 February 2011 (UTC)

There is no royal road to physics. If you really want to understand the subatomic world you will need to learn some math along with it. Said that, I think you do have a point that many of those concepts can be partially explained in plain language but that s not a particularly easy thing to do. Plain language is just inadequate for physics concepts and that is why we use math to begin with. Also, there isn't a bright line separating "known facts" from "theoretical constructs" since the theories are an indispensable component of how physics is understood. Dauto (talk) 16:22, 12 February 2011 (UTC)

consider the first sentence of gluon : "Gluons are elementary particles ..." vs tachyon "A tachyon is a hypothetical subatomic particle ..." and magnetic monopole "A magnetic monopole is a hypothetical ...". Probably, if the article does not mention "hypothetical" in the first few sentences, you may consider it "proven". I suspeci it is the same in biology, where the articles about real animals will not mention they are real, while articles about mythical animals mention they are mythical. 83.134.173.228 (talk) 16:52, 12 February 2011 (UTC)

You should not be surprised that your comments were removed since you added them to the articles themselves. Every article has a talk page associated with it that should be used for comments like yours. Dauto (talk) 17:12, 12 February 2011 (UTC)

Here is a quote from the lead of quark's article:

"Quarks were introduced as parts of an ordering scheme for hadrons, and there was little evidence for their physical existence until deep inelastic scattering experiments at SLAC in 1968.[6][7] All six flavors of quark have since been observed in accelerator experiments; the top quark, first observed at Fermilab in 1995, was the last to be discovered.[5]"

Makes me wonder what are you complaining about?

Did you even read the lead?

Dauto (talk) 19:14, 12 February 2011 (UTC)

It's worth remembering that all serious extensions of the Standard Model (SM) are equally compatible with the experimental facts. There are no known facts that completely rule out supersymmetry or technicolor and yet there is no experimental evidence that we should accept them over the plain old SM. How would you say one is "proven" and the other not? Certainly neither is falsified. 129.234.53.49 (talk) 18:57, 12 February 2011 (UTC)

All of the standard model particles except for the Higgs boson have been experimentally observed. None of the additional particles required by either supersymmetry or technicolor have been experimentally observed. That's the difference. Dauto (talk) 19:09, 12 February 2011 (UTC)

Can you explain what you mean by "experimentally observed"? As far as I'm aware, an extension of the SM with appropriately broken SUSY would have the same low-energy scattering signatures as the SM itself. The fact that at some point the two models would lead to different pheneomena doesn't mean that there is any more experimental evidence to support one over the other. 129.234.53.49 (talk) 20:26, 12 February 2011 (UTC)

I'm not saying SUSY is wrong. But there is no indisputable experimental evidence that it is right either. It is still a hypothesis. For instance, there is no experimental evidence that the selectron actually exists .On the other hand, there is plenty of experimental evidence that the particles of the standard model actually exist. That is a huge difference. Of course that doesn't mean that SUSY is wrong. Dauto (talk) 20:40, 12 February 2011 (UTC)

Well I'm not trying to defend or refute SUSY, that was just an example. My point is that I don't think there is such a thing as "experimental evidence that [some particles] exist". You don't observe proof that something exists, you see experimental results which agree (or disagree) with your theory. Any reasonable extension of the SM will agree with the data gathered so far just as well as the SM does. They are both hypotheses, which are currently not falsified; all current support for the SM is also support for the MSSM and many other theories too. 129.234.53.49 (talk) 23:02, 12 February 2011 (UTC)

Yes, I understand what you're saying but I think the point you're making is a bit disingenuous. The SM is consistent with the data gathered so far because we have gazillions of experiments confirming it. SUSY is consistent with the experiment because we have gazillions of experiments that neither confirm nor refute it. It is not the same thing. There is a difference between being confirmed by experiment and being consistent with experiment. Dauto (talk) 06:02, 13 February 2011 (UTC)

It looks like we may have to agree to differ here. I simply disagree with your last statement; I just can't see how "confirmation" by experiment would work. 129.234.53.49 (talk) 22:35, 13 February 2011 (UTC)

The OP is writing a book, and complaining that Wikipedia's content makes it difficult to evaluate scientific facts. If the original poster is writing a serious book, they should consider hiring a subject-matter expert to review any scientific writing. You may be able to find a local scientist or professor who can contract to review your work for scientific accuracy. You can also contract to various "review shops" on the internet. Physics is difficult; and presenting it correctly and accurately is a challenge. But if any content you wrote were ever challenged, it would be far better to say "...but the content was reviewed for scientific accuracy by Professor So-and-So at Podunk University," rather than "...the content was based on my own interpretation of Wikipedia." A more serious book should be reviewed by somebody with more serious credentials. Physics is an ultra-precise subject, and subtle variations that you may use when wording your presentation of a topic can mean the difference between "completely validated by scientific experiment" and "completely wrong." This applies even to the classical and well-established (non-mind-bending) areas of physics, like Newtonian descriptions of force and energy. "Regular people" can evaluate anything they want for accuracy - but if "regular people" are disinclined from being extremely precise in their thought-process, they will never be able to evaluate physics - no matter how simple and straightforward any presentation may be. Nimur (talk) 20:47, 12 February 2011 (UTC)

I'm not sure whether I'm on the right track here, but the OP seems particularly concerned about counterintuitive notions like spin. The article there explains that spin does not obey the usual rules we expect spinning objects to, so it isn't the same sort of thing that a spinning beach ball has, but perhaps our articles don't make these things terribly explicit. As I understand it, spin is best seen as a mathematical idea that works in quantum mechanics in a way similar to how orbital momentum works for beach balls. The resemblance could stop there, if it should happen to violate intuition, or it could go further, though I myself do not know any more about it. To say that an electron has spin 1/2 is impossible to picture, but it simply means that when you do the math, you find the electron takes a 720 degree rotation (or something mathematically equivalent) in order to reach a state isomorphic to its original one. To say further that it has been demonstrated experimentally only means that we have so much evidence that we have got the mathematics right that we can teach it at undergraduate level without fear of embarrassment. Someone might come up with some deeper mathematics, or a better verbal description, later, but for now we're doing alright. Can a physicist please tell me if I've got this right, and can the OP perhaps tell me if it helps any? It's been emotional (talk) 16:56, 13 February 2011 (UTC)

I certainly agree that Wikipedia has an ongoing problem with math and science articles written by experts, for experts, as opposed to the general population. The sticking point in simplifying such articles is always that the simplified model is not quite as correct. For example, all Newtonian physics is technically "wrong", since it doesn't account for time dilation, etc., and other effects of relativity. So, you can't say that a train going 10 km/hr will travel 10 km in an hour, if any physics PhDs are around, as it all depends on the locations of the observers, etc. StuRat (talk) 04:26, 14 February 2011 (UTC)

Salaries

what is the salary of an engineer working for government after passing IES (Indian Engineering Services exam)? —Preceding unsigned comment added by 1.23.10.106 (talk) 15:13, 12 February 2011 (UTC)

Determining the salary of an entry-level engineer in a government position (I presume it's India that your enquiry is based in) is difficult, as salaries can change from year to year. Additionally, public-sector roles are reorganised regularly, so the job itself may change. Perhaps this article on the Indian Engineering Service, or this link to the Union Public Service Commission of India, might be of use to you. Malusmoriendumest (talk) 08:11, 13 February 2011 (UTC)

Relationship between island size and its freshwater supply

Consider a hypothetical island somewhere in the subtropics. If you want to put a group of say 20 people on the island year-round and want them to be self-sufficient in terms of freshwater needs, how big does the island need to be to have the needed freshwater supply? I know the question is not precisely answerable, but I'd like to see some estimates that has some rational basis. Thanks. --173.49.82.30 (talk) 19:06, 12 February 2011 (UTC)

"Subtropical" describes a zone of temperature not rain fall, so there is no answer to you question. You need to first establish the total water requirements for a 'typical' person multiply by total number of people etc. Err.. just curious: Why do you want to know this – have you discovered the World comes to an end next year and are looking for a bolt hole? --Aspro (talk) 19:26, 12 February 2011 (UTC)

Maybe the OP meant tropical? Near the equator rainfall is regular for most of the year. Here twenty people would not need a large island to meet their water needs assuming they had reasonable equipment for capturing rain water. I can't estimate any kind of specific size, but I think it would suffice to say that in the tropics, water is unlikely to be an issue before food or building materials. --Daniel 19:38, 12 February 2011 (UTC)

Just to add to my earlier thoughts, very small tropical islands usually don't have any naturally occurring fresh water. Rain would be the only source so collection and storage would be very important. --Daniel 19:49, 12 February 2011 (UTC)

The questioner can use the Hawaiian Islands as an example. Size is not as important as elevation. Simply look at Oahu. One side of the island has regular rainfall. The other does not. It is due to the mountains. The wind predominantly blows from the same side. The mountains force the moist wind upward where it cools. Rain forms. For more on this, see orographic precipitation. -- kainaw™ 20:17, 12 February 2011 (UTC)

Another example is Yakushima in Japan. It's relatively small, but also has a very high peak, so clouds bump into the island all the time and it rains almost every day. TomorrowTime (talk) 01:30, 13 February 2011 (UTC)

The most straightforward way to evaluate this is to calculate the size of the drainage basin (which only depends on topography), and average precipitation rate for the drainage basin (which can be easily measured). Excluding ground permeability (that is - water that goes down into the ground, instead of flowing on the surface), this will approximately give you the net surface water outflow rate. This subject can be very complicated - specialists in environmental engineering and water quality management spend years trying to accurately model the water cycle for specific regions.

The next thing to worry about is the nontrivial effect of water-pollution on a very small island. Will humans effectively manage their sewage to guarantee that it doesn't contaminate their drinking water? If so, you need to calculate how much water that would require. Will sewage from livestock, domesticated animals, or wildlife contribute to water-pollution?

Finally, will the humans have any technology (even primitive filtration systems) to clean up polluted or microbe-infested water? If so, they can survive on much smaller quantities of water; otherwise, you will need "fast-moving" streams or rivers - this necessarily requires more water throughput. Nimur (talk) 20:54, 12 February 2011 (UTC)

Freshwater aquifers are important to consider for island water supplies. Smaller and flatter islands will probably have more saltwater intrusion. ~AH1^(TCU) 01:07, 13 February 2011 (UTC)

Odd One out of n Resistors

Hello. There are 10 resistors, nine of which are 1 Ω each. How do I identify the odd resistor and its resistance in the least number of ohmmeter readings? I am allowed zero-resistance wires but no batteries. Thanks in advance. --Mayfare (talk) 21:20, 12 February 2011 (UTC)

Note that the ohmmeter has a battery in it, but I expect that you are allowed to use it. Then, just divide and conquer. If you divide them into two groups of 5 resistors and connect them in series, each should read 5 ohms unless one of the piles has a resistor that is not 1 ohm. Once you identify the pile with the odd resistor, do the same. Divide into two piles and connect all the resistors in the pile in series. Continue until you have only 2 resistors. -- kainaw™ 21:25, 12 February 2011 (UTC)

I think there may be a more efficient way to do it. At the moment this is just a hunch, because I haven't worked out all the details or verified that it actually works, but suppose you do this: First, number all the resistors, so you can distinguish them. Connect the first two resistors in parallel, and then connect that network in series with the third resistor, and then connect that whole thing in parallel to the fourth resistor, and so on, and then measure the equivalent resistance of the resulting network. You can work out what the equivalent resistance should be if all the resistors were 1 ohm; the reading you get will at least tell you whether the odd resistor is higher or lower than 1 ohm. Additionally, you can figure out things like, "If resistor 1 is the odd resistor, then its resistance must be _____; if resistor 2 is the odd resistor, then its resistance must be _____; etc." Now connect the resistors together in a different configuration (I'm being vague here, because I don't know for sure what this configuration should be), and measure the resistance again, and recompute "If resistor 1 is the odd resistor, then its resistance must be _____, etc." Perhaps it is possible to design the two configurations of resistors in such a way that, for any pair of ohmmeter readings, there is only one consistent possibility. If so, then you've solved the problem with just two ohmmeter readings. The details of this idea are left as an exercise to the reader. :-) —Bkell (talk) 21:40, 12 February 2011 (UTC)

A guess: Perhaps the second configuration should be like the first, except that you connect resistors 9 and 10 in parallel, and then connect that network in series with resistor 8, and then connect that network in parallel with resistor 7, and so on. Will that do the trick? It seems plausible to me. —Bkell (talk) 21:49, 12 February 2011 (UTC)

This is a pretty common mind-game or final-exam-question for circuit-design and related EE courses:) Divide-and-conquor is the minimum "student thought about it" answer, based on powers-of-two. There are a million variations/applications of this approach ("how many yes/no questions do you need to ask to determine which playing card someone has picked", etc.). The parallel/series-combinations approach is the really clever solution, solvable in a single measurement for certain numbers of resistors. The question is all over google (both actual science/hobbyist sites and "just tell me the damn answer to my homework question" boardd), but the parallel/series way is only on some of their answers:) DMacks (talk) 21:57, 12 February 2011 (UTC)

Can you really solve it in a single measurement if you don't know the resistance of the odd resistor? I thought about that for a while, and I convinced myself that at least some measurements of the equivalent resistance of the series-parallel network could not lead to a unique conclusion. In particular, if one of the two "innermost" resistors is the odd one out, you have no way of telling which one. —Bkell (talk) 22:02, 12 February 2011 (UTC)

Yes, I'm pretty sure 2 is the best for worst-case scenario or if you need to figure out if the odd one is high vs low. But it can be solved in 1 depending on where the odd one winds up. For example,$-{\begin{bmatrix}-R1-R2-\\-R3-\end{bmatrix}}-$ solves in 2 if the odd one is R1 or R2 as you say, but solves in 1 and even know what the odd one's resistance actually is if R3 is the odd one. Is 4 completely knowable (which R#, what value) in$-{\begin{bmatrix}-{\begin{bmatrix}-R1-R2-\\-R3-\end{bmatrix}}-\\-R4-\end{bmatrix}}-$ by a single measurement unless the odd one is R1 or R2? Is this the start of a pattern where there is always only that one pair that cannot be distinguished by single measurement? DMacks (talk) 22:30, 12 February 2011 (UTC)

Hmm, let's consider that three-resistor network. I agree that for some values of R3, a single measurement is enough to identify it. But if R3 is 1.2 ohms, for example, you will get an equivalent resistance of 0.75 ohm, which is the same equivalent resistance you'd get if R1 was the odd resistor, having a resistance of 2 ohms. So I still think, even if you get lucky and R3 is the odd resistor, you'll sometimes need at least two measurements. —Bkell (talk) 22:54, 12 February 2011 (UTC)

Oooh good point! This is all based on vague recollections of working it out years ago, maybe I was assuming it was known if the problem stated if the odd one was high or low. DMacks (talk) 23:08, 12 February 2011 (UTC)

Well, in the counterexample I just showed, both possible cases have a "high" resistor: the first case has R3 too high at 1.2 ohms, and the second case has R1 too high at 2 ohms. So even knowing whether the odd resistor is too high or too low isn't enough information. —Bkell (talk) 23:18, 12 February 2011 (UTC)

And in your four-resistor network, R3 is in parallel with R4, so if one of those was the odd resistor you couldn't tell them apart. My guess is that you meant to put R4 in series with the three-resistor network. —Bkell (talk) 22:59, 12 February 2011 (UTC)

-R4- is outside the R1/R2/R3 bracket (parallel to the whole 3-resistor circuit. Lemme try again with longer leads...$-{\begin{bmatrix}-{\begin{bmatrix}-R1-R2-\\--R3--\end{bmatrix}}-\\---R4---\end{bmatrix}}-$(wikipedia's TeX doesn't seem to support boxes). DMacks (talk) 23:08, 12 February 2011 (UTC)

Yes, I understand where R4 is. It is still in parallel with R3. The network as a whole has three branches in parallel: one branch consists of R1+R2, the second consists of R3, and the third consists of R4. The network you have drawn is electrically equivalent to $-{\begin{bmatrix}-R1-R2-\\--R3--\\--R4--\end{bmatrix}}-$. The resistors R3 and R4 are perfectly symmetric in this network. I think what you want to draw is $-{\begin{bmatrix}-R1-R2-\\--R3--\end{bmatrix}}-R4-$, where R4 is in series with the three-resistor network. —Bkell (talk) 23:14, 12 February 2011 (UTC)

Ah yeah. There was some pattern for extending each next resistor. And there was also definitely some more specific parameter on the odd-one, since even now you could swap in any of them and get the same result by different choices of the odd-one's value. I'm gonna quit trying to work out the details, been too long since I've forgotten the cleverness:( DMacks (talk) 04:04, 13 February 2011 (UTC)

No two resistors in the network are interchangeable, except for two pairs at the bottom. If all the voltages bridged by the resistors are different, the effect of changing one should be different... I think.

I'm thinking that the circuit I diagram at right should work, provided that you know the unknown resistance, and you're not so unlucky as to put it in one of the spots I failed to make unique (is there a better way to do it? I always end up with some odd resistors out. But I think that if you scramble the resistors and measure twice it should work. But the circuit I draw does bring back fearful memories of the dreaded Wheatstone Bridge from bygone days... and for this to work I'd have to figure out V2, V3, V4, V5... Sigh, let's see, Kirchhoff's circuit laws... Wnt (talk) 06:58, 13 February 2011 (UTC)

Good God, but I'm coming up with stuff like "v5 = (-v1/R1 - v1/R2 +v2/R4 +v2/R1 +v2/R3 +v3/R5+v3/R2+v3/R6-v4/R3)/(1/R4 + 1/R5)". And I still have v4, v3, v2, and v1 to solve for. Wnt (talk) 07:40, 13 February 2011 (UTC)

The kind of network I was suggesting is shown to the right. It is built by connecting resistors D and E in parallel, and then connecting C to that network in series, and then connecting B to that network in parallel, and then connecting A to that network in series; it is built up one resistor at a time by alternating series connections and parallel connections. I believe that no two resistors here are interchangeable except the two innermost ones (D and E). This kind of network has the advantage of being series-parallel, unlike the "grid" of resistors Wnt is building; so you can analyze it using just simple series and parallel reductions, rather than having to perform Y-Δ transforms. I conjecture (though I haven't proved this) that if you build a second network like this, but connect the resistors in the reverse order (swap A and E, swap B and D, C stays where it is) then the measurements of the equivalent resistances of the two networks will identify the odd resistor and its value. —Bkell (talk) 08:02, 13 February 2011 (UTC)

Wouldn't you need 3 readings in the worst case? Say you take 1 reading for the first five resistors. But those resistors are identical. You take another 2 readings to find the odd resistor in the other set of five resistors. I may be confused. --Mayfare (talk) 23:20, 13 February 2011 (UTC)

(To Mayfare:) Oh, yeah, sorry for the confusion. The picture there is only for five resistors; that's only because I had already drawn that picture for a paper I wrote several years ago, so it was easy to just upload what I had rather than drawing a new one. So my explanation was for a puzzle where you just have five resistors. But the idea extends to any number of resistors. If you have ten resistors in all, then just continue that construction: add the next resistor in parallel with these five, and then the sixth resistor in series with that, and so on, adding each resistor one at a time and alternating series and parallel connections. Then find the equivalent resistance of that whole 10-resistor network. Now build the network again, but use the resistors in the reverse order (so the "innermost" resistors from the first network become the "outermost" resistors in the new network, and vice versa), and measure the equivalent resistance of that network. My conjecture is that these two measurements will uniquely identify the odd resistor and its value (ignoring the issues about measurement errors and resistor tolerances that Gr8xoz raises below—I'm assuming that since we are allowed to use zero-resistance wires we also have a perfect ohmmeter and perfect one-ohm resistors). —Bkell (talk) 00:22, 14 February 2011 (UTC)

This method is very sensitive to measurement errors and errors in the resistance of the nine resistors that are suposed to be 1 ohm. 10 resistors connected in this meaner where one of the innermost resistors have a resistance of 2 ohm and the rest are 1 ohm have an total resistance of 0.618055 ohm while if it is the third innermost resistor the resistance becomes 0.618320 ohm. With normal ohm meters and resistors this is easily masked by the errors in measurement and the nine "1 ohm" resistors. I have not analysed the "grid" suggested by Wnt but I would expect it to have similar problems. In order to answer the question we need to know the measurement error, the allowed errors in the resistances and the minimum allowed difference between the odd resistance and 1 ohm. Gr8xoz (talk) 10:57, 13 February 2011 (UTC)

Now you are trying to turn a useless theoretical exercise into something practical. Why would anybody want to do that? Dauto (talk) 15:26, 13 February 2011 (UTC)

February 13

Hill, James B. (1945). Autobiography

An article references "Hill, James B. (1945). Autobiography. Raceland, Louisiana, USA: James B. Hill. pp. 200". How can I get more information on this autobiography? —Preceding unsigned comment added by 66.97.56.130 (talk) 01:34, 13 February 2011 (UTC)

You could try your local library; they're unlikely to have a copy themselve, but they do they have experts in locating books. But it is not certain that it is findable at all. The format of your reference indicates that the book is self-published (the name that comes after the place of publication is that of the publisher), so it may be a small run that has never been catalogued by a general library. The book does not seem to exist in the Library of Congress's on-line catalog. –Henning Makholm (talk) 02:52, 13 February 2011 (UTC)

Also, this would be better asked on the humanities desk. –Henning Makholm (talk) 02:52, 13 February 2011 (UTC)

Time when Sun is highest in the sky

On any given day, is the Sun at its highest in the sky at exactly the same time in all places with the same longitude? (By "same time" I mean same time as measured by a single reference clock. Ignore any issues to do with local time.) I've always assumed, without thinking too much about it, that the answer was "yes", but now I'm beginning to doubt it. 86.183.3.100 (talk) 04:02, 13 February 2011 (UTC)

Why are you doubting it? It seems right to me. Dauto (talk) 04:40, 13 February 2011 (UTC)

Don't doubt it. I've never heard or thought otherwise. HiLo48 (talk) 04:46, 13 February 2011 (UTC)

Yes, local noon has to happen simultaneously on an entire meridian. Noon happens at the moment the abstract plane that contains the Earth's axis and the place in question sweeps through the center of the Sun. But this plane is the same for all places on the meridian, so their noons are the same.

(I'm assuming you're not interested in things like relativistic aberration of the sunlight due to the different rotational speeds on points at different distances from the axis. Not sure the effect is measurable anyway). –Henning Makholm (talk) 04:49, 13 February 2011 (UTC)

One possible exception is in the Arctic or Antarctic, where, depending on the time of year, the Sun may not come up at all that day. So, if there is no Sun at all, the time when it is highest in the sky is undefined (unless, I suppose, you consider that to be when the Sun is closest to, but still below, the horizon). StuRat (talk) 04:10, 14 February 2011 (UTC)

Thanks for the replies. I am trying to calculate the declination of the Sun as seen from Earth at different latitudes, and at different times of year, using a simplified model. The model assumes that the Earth’s orbit is circular and ignores various other complications such as precession. I know that the model is not exact, but it does not seem feasible that making it exact by adding ellipticity and the other tweaks will exactly eliminate the discrepancy that I observe. In the model:

k is number of Earth rotations per year, which I’m assuming is 366.25

alpha is angle of Earth’s tilt, which I’m assuming is 23.5 deg.

a is the time at which we measure the Sun’s declination, in years. a = 0 is the moment of the December solstice in some year (for me it doesn’t matter which one). a = 1 is the moment of the December solstice one year later.

phi is the latitude on Earth from which we observe the Sun

I assume that the point from which we observe the Sun is on the longitude that is exactly facing away from the Sun (i.e. midnight) at a = 0.

I set up a coordinate system whereby the x and y axes lie in the plane of the Earth’s orbit, with the x-axis pointing in the direction from the Sun to the Earth at a = 0, and the y-axis pointing in the direction that the Earth is (instantaneously) moving in its orbit at a = 0. The z-axis points North out of the orbital plane.

To find the declination of the Sun, I first calculate the direction (unit) vector (x,y,z) from the centre of the Earth to the viewpoint on the Earth’s surface at time = a. I do this as follows:

At a = 0, without yet accounting for the Earth’s tilt, we know (x,y,z) = (cos(phi), 0, sin(phi)).

Then, I account for the Earth’s rotation at time = a by rotating vector (x,y,z) about the z-axis (anticlockwise as looking from positive z to negative z) by an angle 2*pi*k*a.

Then I account for the axial tilt by rotating (x,y,z) about the y-axis (clockwise as looking from negative y to positive y) through angle alpha.

Then I calculate theta, the angle between the resulting vector (x,y,z) and the vector from the centre of the Earth to the Sun at time = a, which I make to be (-cos(2*pi*a), -sin(2*pi*a), 0).

Finally I calculate: Sun’s declination = pi/2 - theta

According to my calculations, this results in:

declination = -pi / 2 + acs(cos(alpha)*cos(phi)*cos(2*pi*a)*cos(2*pi*k*a) + cos(phi)*sin(2*pi*a)*sin(2*pi*k*a) + sin(alpha)*sin(phi)*cos(2*pi*a))

The trouble is that I can’t see either numerically or analytically how the maxima of this (w.r.t a) are independent of phi. I get very slight variations.

What am I doing wrong? Thanks to anyone who has got this far! 81.159.104.144 (talk) 14:35, 13 February 2011 (UTC)

Hmmm, actually you are right and we were wrong. I get the same equation as you do.

What went wrong in our intuitive understanding was that we confused "Sun highest in the sky" with "Sun appears due south". They are not exactly equivalent. The "due south" moment (which is simultaneous along an entire meridian) is when the fixed star that the Sun is momentarily in front of has its highest elevation. At that time, this star moves horizontally, but at the same time the Sun is moving along the ecliptic relative to the fixed stars, and this movement generally has a north-south component. So at the "due south" moment, the Sun is actually still moving slightly upwards in the sky, or already moving downwards, depending on which hemisphere we are in, meaning that "highest solar elevation" noon does vary very slightly along the meridian.

This also helps clear up the apparent discontinuity at the poles: When we move closer to the pole, in the limit they daily variation in the Sun's elevation gets much smaller than the variation due to the ecliptic movement. So on any meridian, the "highest elevation" moment during the summer solstice day must converge towards the exact solstice moment as the latitude approaches 90° exactly. –Henning Makholm (talk) 17:50, 13 February 2011 (UTC)

Hi Henning, thank you very much for taking the time to look at this. I'm delighted that you now get the same result as me; staring at these equations has been driving me nuts! 86.179.118.226 (talk) 18:44, 13 February 2011 (UTC)

Henning Makholm wrote" "The "due south" moment (which is simultaneous along an entire meridian)" You are forgetting about the part of the earth south of the Tropic of Capricorn where the noon sun is due north. Between the Tropic of Capricorn and the Tropic of Cancer the sun may be north or south at noon - depending on the date - and twice a year it will be overhead. Roger (talk) 21:15, 14 February 2011 (UTC)

agate spatula

Why is the instrument named agate spatula?? what is the reason 4 it?? — Preceding unsigned comment added by Suprithmurthy (talk • contribs) 04:31, 13 February 2011 (UTC)

It's called an agate spatula because it is made of agate. Apparently these are used by dentists to mix cements, and a metal spatula would be bad because it could react with the cement. Looie496 (talk) 05:01, 13 February 2011 (UTC)

Name for an auction method

My friend described a kind of auction in game theory/economics to me, but I don't know what it's called, and it doesn't seem to be any of the ones described in Auction theory. Here's how it happens:

The seller sets a secret reserve price for the item. Each bidder puts a secret bid in an envelope. The envelopes are opened, and if any bid meets or exceeds the reserve price, the highest bidder wins, but that bidder pays only the reserve price, rather than their bid. If no bid meets the reserve, the item remains unsold.

What is this kind of auction called? —Keenan Pepper 05:42, 13 February 2011 (UTC)

How could this auction work? After the auctioneer has all the bids, he would secretly sneak a peek in the envelopes of a few predicted highest bidders, and set the "reserve price" to just below the highest bid he sees. Alternatively, he'd tell a friend what the reserve price was, and that friend, confident that it was fairly low, could submit an absurdly high bid and clinch the item. Even an honest auction of this type leaves the seller in the peculiar position of accepting less money than the buyers are willing to bid. I have a hard time believing this is actually done - is this a real auction type or just a thought experiment for game theory? Wnt (talk) 06:24, 13 February 2011 (UTC)

Right, it's not a practical method for real auctions. But what is it called? —Keenan Pepper 07:12, 13 February 2011 (UTC)

If it's not practical, then why do you suppose anybody would bother to think of a name for it? –Henning Makholm (talk) 07:23, 13 February 2011 (UTC)

Oh, don't be so naive. Lost of people spend their time thinking about unpractical things. Dauto (talk) 07:44, 13 February 2011 (UTC)

Sure, but things don't usually get named until they've been thought about by several people (who know about the previous thoughts). This is the kind of auction idea that would get thought about once, rejected as impractical, and then not thought of again. --Tango (talk) 21:22, 13 February 2011 (UTC)

It seems to be a model used by game theorists. There is more than one game theorist in this world and sometimes they like talking to each other. Dauto (talk) 23:35, 13 February 2011 (UTC)

Just to be sure...your friend isn't just getting themselves a bit muddled? What they describe sounds quite like a Vickrey auction - except that very last bit...that is to say in a `Vickrey auction' the winning person doesn't pay what they bid, they pay whatever the second highest bid was. Could it be that this is what they were describing? In the auction idea you describe I can't quite figure out what incentive people have to bid low assuming they want to win the item? ny156uk (talk) 09:14, 13 February 2011 (UTC)

Yes, but if they bid a gazillion dollars they may end up having to pay 0.9 gazillion dollars (if that happens to be the secret reserve price) and that might be just a little too much to pay for the used thingamajig he is bidding on. Dauto (talk) 15:23, 13 February 2011 (UTC)

Right, exactly. In fact I believe in this kind of auction the bidders have an incentive to bid their true value, which is why I think it's probably been discussed in a game theory context. —Keenan Pepper 17:48, 13 February 2011 (UTC)

How does this method differ (in a material way) from simply setting a price, telling people what it is and letting someone buy it at that price if they so wish? --Tango (talk) 21:22, 13 February 2011 (UTC)

Don't you think the knowledge of the set price will influence how high people are willing to bid? Dauto (talk) 23:51, 13 February 2011 (UTC)

I agree that this sounds like a second-price auction. As noted above, the equilibrium in this auction is for all players to bid their true valuations (assuming the standard economic utility framework). The more familiar first-price auction provides an incentive for the winning player to over-bid (the "winner's curse.").

As to the auction described, it seems it would be unworkable if more than one bidder valued the item greater than the reserve price. Each bidder will attempt to outbid the other, without constraint (since they only pay the reserve price no matter how high they bid). There is no Nash Equilibrium to this game since the strategy space is infinite and unbounded. It's a little like playing poker for a set number of dollars per hand and an infinite bankroll; the strategic possibilities would break down very quickly. —Preceding unsigned comment added by 12.186.80.1 (talk) 16:07, 15 February 2011 (UTC)

Smallpox Vaccine

What is the big needle used for when giving a smallpox vaccine?

Smallpox Vaccine

Vaccine Injection

Also, why do people who have gotten the vaccine have a scar on their shoulder? Scar on Shoulder —Preceding unsigned comment added by 76.169.33.234 (talk) 07:53, 13 February 2011 (UTC)

Sorry - I took the liberty to change your figures over to the standard thumbnail format, because they were too big and foul up display of the Refdesk. I didn't know that by using a single bracket http link to the thumbnail you can get them to display, but the humdrum way is more practical. Wnt (talk) 07:59, 13 February 2011 (UTC)

Thanks! I couldn't figure it out, first it was too big, then it was just links. Looks much better now. —Preceding unsigned comment added by 76.169.33.234 (talk) 08:01, 13 February 2011 (UTC)

You may be interested in Smallpox vaccine#Post-eradication vaccination Nil Einne (talk) 08:17, 13 February 2011 (UTC)

The reason for the needle's shape and how its used is given in the article Bifurcated needle.--Aspro (talk) 11:23, 13 February 2011 (UTC)

Thanks for the info. The article states that the established method was to do it in the left arm, was there any particular reason for this? Why not do it someplace else where the scar that is left is not visible? Is the permanent scar that is left because of the vaccine or because of the needle? —Preceding unsigned comment added by 76.169.33.234 (talk) 11:53, 13 February 2011 (UTC)

The scar results because the vaccine is a live virus, which multiplies at the site of introduction (in the cells around the skin puncture). One's immune system then kicks in and confers immunity which prevents the virus spreading and causing more 'pox marks.' This local infection is what causes a scab to form, which eventual drops off. The needle just needs to puncture the skin enough to see a little blood. This is because unbroken skin is a very good barrier and may prevent the virus from successfully invading the live skin cells below. The tiny puncher marks on their own, would otherwise heal without a trace. So it is the infection that leaves the scare.

People who have had small pox, very often carry pox marks for the rest of their lives.

The inoculation is performed on the opposite side to the dominant hand because as it is a live virus and a real irritating infection, the site is commonly tender and sore -and can itch like mad. In a few people, it can ache so much that they like to have the arm in a sling. Hence the preference for the arm on the opposite side. It is for this reason too that it is better on the arm than the leg. Also, there is the practical aspect that in an inoculation drive, it is more convenient to have people (maybe a thousand or more in a day) form a line with their sleeves rolled up. One doesn't have to lean too far down to reach each arm. You never find medical students picking grapes on their vaccation – too much bending!--Aspro (talk) 15:32, 13 February 2011 (UTC)

If the smallpox vaccine were still given today routinely like it was in the past, do doctors now have something that they can give to the patients to minimize the scars? --173.49.14.8 (talk) 04:30, 14 February 2011 (UTC)

In the past the scars were not considered an issue because almost everybody had one, so I would image they would be dismissed in the same way today. The only treatment I can think of, that may reduce scaring, would be to stop the blister from drying out and forming a hard scab. This could be done by applying silicone scar sheets and then continuing for some time after the dead skin sloughs off, with the topical application of a suitable skin gel. Other treatments such as ex-foliation of the skin etc., to encourage it to regrow smooth is I think – vanity going too far.--Aspro (talk) 11:37, 14 February 2011 (UTC)

Note that the smallpox vaccine isn't the only vaccine which produces a scar. The Bacillus Calmette-Guérin is also know for the scarring although that is usually applied to the upper arm and with a more normal hypodermic needle (at least it was in Malaysia) AFAIK. The BCG uses attenuated live bacteria however a live vaccine obviously doesn't guarantee prominent/significant scarring as it depends on specific vaccine and the bodies response thereof, for example I don't believe the MMR vaccine is notable for any particular scarring. Nil Einne (talk) 07:24, 14 February 2011 (UTC)

Vasectomy: Potential post-operative hormonal changes

The standard article on vasectomy states that after the procedure, "Sperm are still produced by the testicles, but they are broken down and absorbed by the body." Several external articles (which may or may not be objective) put the "average time" for spermatozoa to be reabsorbed into the body after production at around 40 days (but their cited source is obscure). Many papers are available on vasectomy itself, post-operative complications and chronic pain, but I am unable to determine whether any studies have been conducted on post-operative hormonal changes. My question pertains to whether male hormonal levels (specifically those of androgens such as androstenedione and testosterone) are likely to increase measurably after vasectomy, given that sperm cells and their constituent elements are no longer being expelled from the body, but broken down and re-used. Any study or article that shows measurable change (of even the smallest degree) might be relevant; failing that, even anecdotal evidence might be of use, provided that its evidential value is reinforced by multiple instances. Consulting urological surgeons has not yielded results, since their speciality might be said to lie in the performance of the procedure, not its effect on the delicate balances of blood chemistry.

Malusmoriendumest (talk) 08:05, 13 February 2011 (UTC)

All I can give is my opinion. I would say no, it shouldn't have any affect on hormone levels, since breaking down of cells, such as sperm, would involve them dying, being disassembled into base components, then reassembled into other things or excreted. StuRat (talk) 04:02, 14 February 2011 (UTC)

Temperature and wavelengths

Is it possible to contain all wavelengths/frequencies at a specific temperature? For example does Sun emit frequencies in the range of 0 to ∞ Hz, and does our body do the same thing?--Email4mobile (talk) 08:33, 13 February 2011 (UTC)

blackbody radiation 83.134.173.228 (talk) 10:05, 13 February 2011 (UTC)

I think I couldn't find a clear answer in that article. Let me give a direct example: Is heating iron to 1000 °C associated with Gamma rays emissions or Radio frequency waves?--Email4mobile (talk) 10:18, 13 February 2011 (UTC)

I was reluctant to say anything more because I don't want to risk saying anything false about extremely high energy photons. Anywaw, as far as black body radiation applies, any body at nonzero temperature emits radiation at all wavelengths, so the answer to your original question is yes (you personally emit some extremely small amount of gamma rays). Nevertheles, most of the radiated energy goes into radiation at a certain frequency range that is a function of the body's temperature. So 1000°C emits mostly visible light and infrared, but it also emits some negligibly small amount of radio waves and gamma rays. 83.134.173.228 (talk) 11:07, 13 February 2011 (UTC)

The amount of radio waves is not negligible. Dauto (talk) 15:06, 13 February 2011 (UTC)

The article on Planck's law shows the rather slewed bell curve of the spectral output verses temperature. It is the peak of the curve that corresponds to a specific temperature.--Aspro (talk) 11:37, 13 February 2011 (UTC)

I haven't worked out the numbers for when this would happen, but if you have a cold enough object and you are interested in a short enough wavelength then, while technically some radiation at that wavelength will be emitted by that object, it may work out as less than one photon during the entire age of the universe. That can be interpretted as essentially not emitting anything at that wavelenght. --Tango (talk) 21:31, 13 February 2011 (UTC)

I was interested in this question, and my back-of-envelope calculation suggests that for normal-sized black-body objects at around room temperature, this cutoff point, beyond which a total (over all wavelengths) of less than one photon is emitted during the entire age of the universe so far, lies somewhere in the visible spectrum. (Don't rely on this answer though ... I could have done something drastically wrong!) 86.160.216.197 (talk) 21:40, 14 February 2011 (UTC)

which software

please any experienced person help me .. which software i should i use to model a disintegration of nuclear fuel rod(fission process).please give any other useful regarding its modelling —Preceding unsigned comment added by 59.93.130.74 (talk) 09:33, 13 February 2011 (UTC)

There are different Nuclear fuel cycles. This article may help you focus in on the one that suits your application best. Then you can start looking for the software that models that cycle. Thorium is thought to be the next big thing in reactor design. --Aspro (talk) 11:42, 13 February 2011 (UTC)

Perhaps User:Rocketshiporion can help? Their interest seemed to be in the nuclear weapon side but they may have picked up something in their research that may be of use Nil Einne (talk) 16:08, 14 February 2011 (UTC)

physics

what is horizon of a time machine? — Preceding unsigned comment added by Sanandnps (talk • contribs) 14:38, 13 February 2011 (UTC)

You can't title your question "Physics" and then ask about time machines. Since time machines only exist in fiction, the entertainment forum might be a better place for your question. Dauto (talk) 15:46, 13 February 2011 (UTC)

Why answer so rudely? 1) Time travel is a very real topic in physics, and the time travel article has dozens of references from peer-reviewed physics journal articles, 2) if you don't have an answer, why post? — Sam 67.186.134.236 (talk) 18:17, 13 February 2011 (UTC)

I didn't mean to be rude., I think my answer is the best answer to the question. Dauto (talk) 20:05, 13 February 2011 (UTC)

That depends. Is the hypothesized time machine traveling forward or backward in time? If traveling forward in time (faster than normal), one way to do that would be to hang out just outside the event horizon of a black hole, as mentioned in Time travel#Time travel to the future in physics. Traveling back in time may or may not be possible, depending on whether closed timelike curves actually exist. The boundary of a set of closed timelike curves is a Cauchy horizon, which is touched on lightly in Time travel#Other approaches based on general relativity. Red Act (talk) 16:21, 13 February 2011 (UTC)

By the way, even if travel along a closed timelike curve is possible (and there are some very strong constraints on that possibility, such as Hawking's constraint prohibiting a compactly generated Cauchy horizon in a region where the weak energy condition is satisfied), the nature of doing that would be very different from the type of time travel that's pretty much always depicted in fiction, in which the time machine or time traveler has a discontinuous world line. A discontinuous jump in a world line like that is just fiction. Red Act (talk) 17:38, 13 February 2011 (UTC)

Dang, I just wish for the days when all you had to do was push this little button and you appeared in a futuristic city filled with Eloi. Crimsonraptor | (Contact me) Dumpster dive if you must 18:46, 13 February 2011 (UTC)

Its horizon seems to be only a few seconds - yours, a dissatisfied customer.

Dear Sir: I am writing to complain about the Mark I time machine that you sold me ... Gandalf61 (talk) 13:36, 14 February 2011 (UTC)

Pig feed

What are domesticated pigs at pig farms fed? (historical info would be interesting as well)Mintyf (talk) 15:41, 13 February 2011 (UTC)

It's unsourced and not terribly detailed, but we do have some information in the second paragraph of Intensive_pig_farming#Intensive_piggeries and at Sty#Slopping_the_Hogs. --Allen (talk) 18:24, 13 February 2011 (UTC)

They're fed pelletized grain. --Sean 15:38, 14 February 2011 (UTC)

For historical information, Toilets_in_Japan#History states that:In Okinawa, the toilet was often attached to the pig pen, and the pigs were fed with the human waste product. This practice was banned as unhygienic after World War II by the American authorities....though I don't believe that practice was widespread.Smallman12q (talk) 19:28, 14 February 2011 (UTC)

What is this plant?

http://img.skitch.com/20110213-mykikb758gb8g8si9q5kh865gi.jpg

Hi all,

Does anyone know what this plant is? We were thinking it was a Wandering Jew, but now I don't think it looks that much like one. Thanks so much! — Sam 67.186.134.236 (talk) 18:07, 13 February 2011 (UTC)

I know it as a Wandering Jew too, but I think it's one of the Tradescantia family. --TammyMoet (talk) 18:28, 13 February 2011 (UTC)

Maybe specifically Tradescantia spathacea—cf. this Flickr image. Deor (talk) 18:34, 13 February 2011 (UTC)

Thanks! — Sam 67.186.134.236 (talk) 21:15, 13 February 2011 (UTC)

Low-density solids/liquids?

Are there any solids or liquids that are less dense than air? --75.15.161.185 (talk) 21:01, 13 February 2011 (UTC)

Our article Aerogel has some information on this topic:

"The world's lowest-density solid is a silica nanofoam at 1 mg/cm³, which is the evacuated version of the record-aerogel of 1.9 mg/cm³. The density of air is 1.2 mg/cm³."

So it seems there are solids lighter than air, but they're highly porous materials.

Ben (talk) 21:09, 13 February 2011 (UTC)

So why don't they float off into space? --75.15.161.185 (talk) 21:16, 13 February 2011 (UTC)

I think "evacuated version" means what you get when you put it in a vacuum. In air, it has a density higher than air so doesn't float away. In a vacuum, there is nothing for it to float in. --Tango (talk) 21:34, 13 February 2011 (UTC)

so then in other words, because the structure of an aerogel is always "thin strands of open-cell structure of something, with air inbetween," the answer is no? 65.29.47.55 (talk) 22:58, 13 February 2011 (UTC)

So are there any solids or liquids that are less dense than air when in the atmosphere (i.e. not in a vacuum)? --75.15.161.185 (talk) 23:01, 13 February 2011 (UTC)

I think that you should relax your standards to allow for any gas, and it will be hard to find such a case anyway. If you do this, then a gas like sulfur hexafluoride (or for those willing to put up with nasty nasty HF, tungsten hexafluoride) can be up to ten times denser.

Now an aerogel is not "truly" solid throughout; it's a sort of mass of fibers. If you allow something not solid throughout, you might as well allow a boat. Which brings us to [3], which is about as close as I think you're going to get to what you want. Wnt (talk) 00:01, 14 February 2011 (UTC)

Instead of going with another gas, you could go with more pressure. This would compress air more than most liquids or solids, thus making them more likely to float. StuRat (talk) 03:55, 14 February 2011 (UTC)

I think the question is nonsensical, imagine asking the opposite: is there a gas more dense then solid? Obviously not; if it was more dense, then it would BE solid. Vespine (talk) 04:52, 14 February 2011 (UTC)

I don't think that follows: for example liquid Mercury is much more dense than solid Sodium at room temperature. AndrewWTaylor (talk) 09:25, 14 February 2011 (UTC)

The question and most of the discussion is about "solids and liquids" vs gases, not "solids vs liquids". Solids and liquids generally do have similar densities (molecules just less organized and a little less tightly attached to each other in liquids), whereas gases are generally orders of magnitude less dense (molecules highly spaced) at normal pressures and temperatures. DMacks (talk) 16:44, 14 February 2011 (UTC)

One thing worth noting is that the definition of a "solid" is a wee bit unclear. The old rule-of-thumb "conforms to the shape of its container" (like water becomes a cylinder when poured into a glass) becomes a bit weird for gels, infusions, and glasses (silica glass has a barely-measurable viscosity). Molecularly, one might call a solid as being a stable crystal lattice, of which there are many configurations like the sorts in the phase diagram of water/ice, while polymers and glasses would be amorphous solids (or, in physics, soft condensed matter).

The point is that aerogel is a solid as most would think of it, while a highly porous object, like the very light pumice rock, is cheating a bit (especially if we filled the holes with helium and sent it up like a zeppelin. But if you want a nice crystal structure, then here's a fun exercise: check out the [formula for theoretical density of a crystal] and compare it to the ideal gas law to see if you can get your very low-density configuration of some light atom or molecule - the point of the ideal gas law is it sets an upper limit on your solid's volume. SamuelRiv (talk) 23:39, 14 February 2011 (UTC)

For a theoretical solid you could try solid dipositronium, unfortunately it decays too fast to be realised, but it would be lighter than air and solid, but at cryogenic temperatures. Graeme Bartlett (talk) 23:52, 14 February 2011 (UTC)

February 14

Sulfur hexafluoride boat

If you had a large enough tank filled with SF₆, could a boat with a person in it float on it? Would you be able to move by paddling with the oars? --75.15.161.185 (talk) 00:19, 14 February 2011 (UTC)

Yes and yes, though the scales would need to be rather large and your person would need breathing gear for safety. For the latter part particularly, consider the general nature of propellor-driven aircraft. — Lomn 00:41, 14 February 2011 (UTC)

You'll need much more gas than Mythbusters used. Clarityfiend (talk) 00:52, 14 February 2011 (UTC)

From the Sulfur hexafluoride, we have an SF₆ density of 6.12 g/L or 6.12 kg/m³. From Density of air we have 1.29 kg/m³. The difference gives us a net of 4.83 kg/m³ buoyancy. If we needed to float 50kg (say a 40kg person in a 10kg boat), the boat would need to displace about 10.4 m³. Contrast with water displacement, where only about 0.05 m³ (50 liters) need be displaced to float that same 50kg. -- Tom N (tcncv) talk/contrib 01:10, 14 February 2011 (UTC)

I would just mention that, while the videos look very cool, sulfur hexafluoride is a very powerful greenhouse gas. I'd love to try the thing where you breathe SF₆ and talk (it's the flip side of the helium thing) but I'm not sure I can justify it ethically. --Trovatore (talk) 01:13, 14 February 2011 (UTC)

Wow, you're not kidding! 23,900 times more efficient than CO2 on a 100-year time scale, 32,600 times more effective on a 500-year time scale. (Though at least that's not by weight!, but by parts per billion, which according to Parts-per_notation#Air_measurements seems to be a volume-based measurement.) My, I wonder if you can find sulfur and fluorine on Mars...^[4]

In terms of size, this would be more like and airship than a boat. The volume required to lift a person would still be much lower than helium baloons in air, though. — Preceding unsigned comment added by Roberto75780 (talk • contribs) 12:41, 14 February 2011 (UTC)

Youtube video - mechanical glitch, other artifact or mysterious unreflective object?

Hi. I recently came across this Youtube video showing images of a dark object detected by STEREO but according to an astronomer referred to in the comments is not visible on LASCO. I would be interested to know what it is. I realize the video description and annotated text could be exaggerated and jumping to only one conclusion but can a glitch persist for a month and change in size? Thanks. ~AH1^(TCU) 00:21, 14 February 2011 (UTC)

If you want to be taken even remotely seriously, do not get your STEREO data from Youtube.com. NASA makes the STEREO data available to the public at the official spacecraft webpage, STEREO, hosted at Goddard Space Flight Center. You can download the official data. If a "glitch", image-processing artifact, or astronomical observation of any significance were present in the data, it would not be reported through pseudoscience-crank-videos on Youtube.

Although, if NASA was in on the conspiracy, they might have distorted the images in the official database. We can only presume that if it is the case, and NASA has covered up these glitches by doctoring the official data, that the author of the Youtube video was able to independently acquire the unadulterated raw video, with glitches intact, by operating his own Deep Space Network-sized ground stations; and that he has the scientific expertise to interpret and analyze the images and ascertain the presence of a mystery-object. As an anonymous Youtube-author, though, the uploader of that video has a bit less credibility than NASA.

If you are actually interested in the scientific explanation for image artifacts, NASA's official page has an entire set of explanations for each type of artifact on each instrument on the STEREO spacecraft. Image Artifacts, from NASA, with various subpages.

Here are some nice examples of internal reflections of Venus. These sorts of optical artifacts are very common. If you've ever photographed through a fancy compound optics system, you know how much of a problem internal refractions/reflections/distortions can be. I can personally attest to some funny optical trickery back in December while I was photographing Venus in the early morning. I could have sworn I saw a spurious bright spot - a "UFO," or a moon, hovering just off the limb of Venus. So here was this bright shiny object that I could see, plain as day through my eyepiece, but was completely immune to being photographed! Magic! Or perhaps alien technology! What actually had happened was that my camera, having been sitting right next to my warm cup of coffee, had developed some "fog" or dew on one of its internal optical surfaces (probably the mirror) while I changed lenses; and I got a very nice bright spot that was showing up only through my eyepiece (and never in my photographs). After a few minutes of thermal acclimation, the foggy layer dissipated and my "UFO" disappeared (and I was unable to claim discovery for a moon of Venus). Nimur (talk) 00:58, 14 February 2011 (UTC)

(The specific "glitch" reported in the Youtube video was an example of the long-exposure noise reduction algorithm - it is a digital postprocessing artifact. This effect only shows up on certain data-products. Scientists studying STEREO data will usually obtain low-level data products (the rough equivalent of "shooting raw images") and apply a smarter noise-reduction method. I apologize if my earlier anecdote implied that the "dark spot" was some type of optical effect - I only meant to emphasize that astrophotography is full of "spurious data." Nimur (talk) 13:48, 14 February 2011 (UTC)

Nimur, that is an outstanding response. Thank you for taking the time to write it, and I especially enjoyed the artifact links you included. You are a credit to your species. The Masked Booby (talk) 00:58, 15 February 2011 (UTC)

Antibacterial gels vs iodine

Hi all,

Is there any difference between using antibacterial/antiseptic gels such as Neosporin vs. old-fashioned iodine (or even rubbing alcohol)? Is one better in some cases than in others? Thanks! — Sam 63.138.152.135 (talk) 14:49, 14 February 2011 (UTC)

Non-iodine is better in cases where the patient is allergic/sensitive to iodine. --Sean 15:45, 14 February 2011 (UTC)

And conversely for patients allergic/sensitive to neomycin. Iodine in general can some adverse effects (scarring and interference in healing) depending on formulation. Alcohol evaporates and then it's gone with no lasting effect. Some pathogens are particularly sensitive or insensitive (or worse, develop resistance) to specific agents (especially because some agents are broad-spectrum and others are more mechanism-specific in their mode of action). Hard to answer the question specifically because "in some cases" is pretty vague...includes everything from playground scrapes to surgical incisions to cases of MRSA. DMacks (talk) 18:15, 14 February 2011 (UTC)

It depends a bit on why you're using it. The modern version of good ol' fashioned iodine is Povidone. One important difference between the two is that if you where to use an iodine preparation on a daily basis, one would eventually absorb too much through the skin. However, for some indications it is certainly better. In the first world war, alternatives to iodine was garlic and cannabis --Aspro (talk) 20:21, 14 February 2011 (UTC)

Neosporin also contains a skin protectant and the antibiotic in it is benign, it doesn't cause cell death. The AMA is no longer recommending you use alcohol or peroxide on wounds (not sure about iodine I know it's used in medicine still for example prior to catheterization in a female), the fact it kills cells means that it does more harm than good. 65.29.47.55 (talk) 22:31, 14 February 2011 (UTC)

Atomic structure

When one atom moved away from anywhere very time atomic level are not same but it more equilibrium than other for fixed position with some wave not particles very much to minimum ways series.Reaction position are not same means the wave are not similar for bad condition create some how. — Preceding unsigned comment added by Vowies (talk • contribs) 17:53, 14 February 2011 (UTC)

I'm sorry, but don't think I understand you. Can you try writing that again using proper grammar and punctuation? Dauto (talk) 18:59, 14 February 2011 (UTC)

Or, if English isn't your first language, please write in your first language and someone will translate. Or, see if there is a reference desk on the Wikipedia in your first language. --Tango (talk) 23:17, 14 February 2011 (UTC)

The Ever Eluding Question

Do plants feel pain?? Dutch scientist Marcel Dicke, of the Agricultural University in Wageningen, Holland, found evidence that all plants perform similar actions to the trees, when under threat from predators. Indeed, the level of sophistication in this process is made all the more remarkable by the fact that the these ‘signals’ encourage production of substances tailored to specific pests! An example of this would be the lima bean. When attacked by spider mites, the plant releases a chemical attractant for other types of mite, which prey on the attackers. Some plants help others, as in the case of cabbages, which release foul smelling isothiocyanates, discouraging aphids from attacking neighbouring plants like broad beans. —Preceding unsigned comment added by 1.23.10.106 (talk) 18:14, 14 February 2011 (UTC)

I agree these are survival mechanisms and nowhere close to "animal" like senses but how to concisely refute the theory that plants feel pain in? —Preceding unsigned comment added by 1.23.10.106 (talk) 18:19, 14 February 2011 (UTC)

Plants do not have pain nerves or a brain to receive signals from the pain nerves. To claim that plants feel pain will require defining what you mean by pain in such a way that it is not dependent on the biological process of receiving and processing signals from pain nerves. -- kainaw™ 18:36, 14 February 2011 (UTC)

Nociception may be one such term - for example, Paramecium displays nociception, trying to back off and evade a fine needle.^[5]

One way in which Temple Grandin concludes that many (not sure about all) insects do not feel pain is that they will walk on broken limbs, which no vertebrate will do. Just putting that out there. From Animals in Translation. --Mr.98 (talk) 18:58, 14 February 2011 (UTC)

Here's a related question: how do you concisely show that human beings (other than yourself) feel pain? It's entirely possible that you're the only sentient human, and that all other humans are robots without any consciousness. If you punch a human, how would you know whether that human is responding because he has the subjective experience of "pain", or because he's hard-wired to respond to harmful stimuli so as to minimize that stimuli.

In this case, you might use induction to argue that because you feel pain, and because other humans have nearly identical biochemistry, organs, nervous systems, and reactions to stimuli, they probably feel pain too. However, you can't conclude from this that plants don't feel pain, because you can't prove that the mammalian body is the only system that can feel pain. In fact, it would be very surprising if this were true: why should this one path that evolution semi-randomly decided to take be the only one that leads to the subjective experience of pain? --99.237.234.245 (talk) 20:20, 14 February 2011 (UTC)

As stated above - you need to define what it means to "feel pain" as you are purposely not using the biological definition. -- kainaw™ 20:43, 14 February 2011 (UTC)

Pain: the conscious, subjective experience that something unpleasant or harmful has happened. I think this definition is much closer to what people usually mean by "pain". Since the OP obviously isn't asking how to prove that plants have no nerves or brains, I assumed that he meant "pain" in the everyday sense of the word, and was not using the biological definition. --99.237.234.245 (talk) 23:03, 14 February 2011 (UTC)

I don't think plants can be considered to do anything "consciously". That would require a central nervous system. --Tango (talk) 23:22, 14 February 2011 (UTC)

There is no evidence that a central nervous system is required to be conscious. The fact that humans have a central nervous system means only that the one evolutionary path Earth's animals happened to follow can lead to consciousness. It does not imply that this is the only path, or the most likely path, or the most common path. For all we know, sentient alien creatures might have an entirely chemical "nervous system", transmitting chemicals along structures similar to Earth plants' xylem and phloem. --99.237.234.245 (talk) 01:15, 15 February 2011 (UTC)

Not to detract from the scientific interest of the question, but maybe you might ask for modern philosophy references from the humanities desk. The reason I suggest this is that beyond what we know about nervous systems and empirical science (and I think Temple Grandin's comment is an excellent empirical argument), there is an existential element of the question to which our backgrounds might not do justice. To that end, check out Bentham's animal rights philosophy, and this nice response with lots of links (but from an MA only, so take appropriate salt). And of course our paranormal plant perception article goes into the actual arguments a bit as well. SamuelRiv (talk) 00:03, 15 February 2011 (UTC)

I've read that some trees produce a toxic substance when animals eat their leaves. This prompts the animals to move to the next tree. You could perhaps consider this as an algorithm executed by the tree to protect itself. Arguably, pain is the running of any type of algorithm that overrides/restricts normal function. Count Iblis (talk) 01:02, 15 February 2011 (UTC)

You jostle a car and an alarm sounds. It is an algorithm, perhaps a basic sort of nociception, but does the car feel pain? In truth we have nothing resembling a theory of consciousness. We say that certain neural pathways carry pain because individual persons say they don't feel it when they are anaesthetized or severed. But how do you know even that what a person feels is only what they say? How do you know that the person who comes out of the operation in twilight sleep isn't saying he's alright, but somewhere "unconsciously" he has been in agony? Can you generalize further to say that people are the only consciousness, and no other pathway can feel pain? I don't see a scientific basis to disprove that the clouds feel pain when an airplane flies through them. Wnt (talk) 15:22, 15 February 2011 (UTC)

That is the problem here. In order to claim that plants feel pain, the definition of pain is being generalized to purposely include plants. The side-effect of that is that other things (like cars) get included in the generalized definition. So, it is not a discussion about feeling pain. It is a semantic argument about how to purposely include plants without including other things - which is not in any way science. -- kainaw™ 15:26, 15 February 2011 (UTC)

I agree, but then one can always assume that a conscious entity is identified with an algorithm that is running. This is equivalent to the strong AI hypothesis. So, Wnt is generated whenever we run that particular program that is currently implemented by the neural network that his brain is running. Any entity that is experiencing pain clearly has its normal algorithm being interfered with by the running of an emergency algorithm. In case of a car, you have to assume that without the alarm going off, there is an algorithm running and from the perspective of that algorithm, the alarm going off, is restricting things in some way. Of course, this approach can be criticised as being not very scientific. However, it may be that deep down all that exists are algorithms, i.e. that all that exists is simply a mathematical multiverse. It then makes sense to talk about the world that any given algorithm subjectively perceives as a mathematical representation of the algorithm. Count Iblis (talk)

You are touching on the philosophical question of "Other Minds." There has a rich tradition going back at least to Descartes. A good introduction is the Brain in a vat article. —Preceding unsigned comment added by 12.186.80.1 (talk) 16:19, 15 February 2011 (UTC)

Time dilation measured by radioactive decay

I have a question concerning my better understanding about time dilation. I am wnadering if it is just mathematic aspects or it does really have physical meaning.

Supose the experiment with two observers. One in land, stationary with no movement. The other is travelling inside one rocket with speed around 0.8 light speed (c). If they have the same amount of radioactive matter, uranium or others. What we would observe when the rocket return to start point after one travel that had taken 01 thousand years. Would they see different carbon contents in these two samples or it will be the same. My question is based in the fact that time dilation always has been explained using the time interval between events. This trial has different concept.

Obs: I am not english native. — Preceding unsigned comment added by Futurengineer (talk • contribs) 20:18, 14 February 2011 (UTC)

This experiment doesn't have a different concept. If the time interval between individual particle decays dilates, the half-life must get longer, because the sample is now decaying more slowly. If you put uranium on a spaceship and make it travel at high speed, the spaceship sample will decay more slowly as seen from Earth than an Earth-bound sample, by a factor of gamma=1.67. --99.237.234.245 (talk) 20:36, 14 February 2011 (UTC)

The basic answer is yes, it has real physical meaning. It can be measured a number of ways to prove that this is the case. It is not just a mathematical construction. --Mr.98 (talk) 20:54, 14 February 2011 (UTC)

And to answer your question more directly: There will be more radioactive material left over in the spacecraft sample (as compared to the earth's sample) because a smaller amount of time has elapsed in the spacecraft than on earth. Dauto (talk) 21:43, 14 February 2011 (UTC)

For example, in introductory physics, a common homework problem requires application of relativistic time dilation to explain the arrival rate of muons originating from cosmic rays. These muons decay too quickly, and they have no business arriving at Earth's surface as frequently as they do; but because they are traveling at relativistic speeds, the decay half-life measured in the laboratory is not the same as the decay half-life measured in the co-moving frame of the muon. A model of cosmic ray production in the upper atmosphere, from SLAC, works the math properly for you, and explains the physics of muon production as a result of cosmic hydrogen ions colliding in the thermosphere. Nimur (talk) 23:25, 14 February 2011 (UTC)

Magnetic confinement fusion

After reading this article, I still have no idea how it's supposed to get fusion to occur. Is the idea to get hot plasma as dense as possible and loiter it around each other until the particles can quantum tunnel through and produce fusion? ScienceApe (talk) 21:23, 14 February 2011 (UTC)

Fusion is actually the trivial part: heat hydrogen enough and it just happens. The hard part is making it continue to happen: as soon as your fusing plasma touches any normal-temperature material, its heat is sapped away and the reaction stops; similarly if it spreads out too much. Thus "confinement"; magnetic confinement is just the clever application of magnetic fields to hold the fusing plasma in place for a "sufficient" time to be interesting/productive. --Tardis (talk) 21:40, 14 February 2011 (UTC)

Couple of questions. How do they heat it in magnetic confinement? I know they use lasers in inertial confinement (usually). Why is the torus shape preferable over a spherical shape? Would this be easier to do in a zero gravity environment like outerspace? ScienceApe (talk) 21:45, 14 February 2011 (UTC)

I think it's usually RF heating: the plasma is conductive and can receive energy like an antenna. Torii are good because the magnetic field is solenoidal: the field lines have to form loops, so it makes sense to make the device a loop as well. I don't think gravity is a problem: one ought to be able to counteract it with a static electric field (and a slightly charged plasma). --Tardis (talk) 21:54, 14 February 2011 (UTC)

Watson the computer system

Would the Watson computer system likely be capable of passing the Turing test? Googlemeister (talk) 21:46, 14 February 2011 (UTC)

No. Dauto (talk) 22:01, 14 February 2011 (UTC)

(EC) Presuming you mean Watson (artificial intelligence software), that seems rather unlikely. Amongst other things, if it could you would think IBM would be making more noise about that rather then hyping its Jeopardy! skills... Nil Einne (talk) 22:05, 14 February 2011 (UTC)

IBM isn't even claiming that their system is "turing-test ready." They're simply advertising that it has sophisticated capability for answering questions (or ... questioning answers) posed in natural language format. (I'll also comment that those East-Coast intellectuals seem to have designed a nerdy operating system. Out here in California, our robots are free-range outdoorsy-types who enjoy hiking and long walks on the beach... not crashing out in front of the TV watching Jeopardy. Nimur (talk) 22:33, 14 February 2011 (UTC)

The article you linked has a section on the [Prize], for which "the first contest was won by a mindless program with no identifiable intelligence that managed to fool naive interrogators into making the wrong identification," which is particularly evident in Turing competitions where they have ordinary citizens as judges. But for some perspective on what the Turing Test actually means in terms of consciousness, maybe check out both recent and early transcripts of the Loebner competitions. For all of Turing's genius, his test appears to be short-sighted - perhaps we should start thinking about an improved version video? SamuelRiv (talk) 00:39, 15 February 2011 (UTC)

There is no "official" Turing Test, so the question is unfortunately a little too vague. The Loebner Prize above has tried to make itself into the "official" test, but it's not a test that any scientists participate in, only chat-bot writers with lots of time on their hands. (And they are still pretty terrible -- here's the transcript from the 2009 winner. Really no different from the early Eliza programs.)

One answer might be "Does Watson play Jeopardy in a human-like manner?" That is, might you think a human was giving those answers? If so, Watson has passed a very limited Turing Test. Watching the game last night, the answers were very good, unlike a few months ago when the answers didn't match the questions. So it may have passed this Turing Test. In terms of conversational ability, however, it's not even trying. Therefore, it wouldn't pass the Turing Test as normally envisioned. — Sam 63.138.152.135 (talk) 14:43, 15 February 2011 (UTC)

There is already a method which can answer questions given sufficient information in the query and the database. An example would be if you asked, "What country is known by a pattern of 13 alternating red and white stripes with 50 white stars on a blue background in the upper left hand corner or similar question?" (this example) --Inning (talk) 17:43, 15 February 2011 (UTC)

February 15

California power plant's NOx emissions

I have had a good look around and have been unable to find an answer for this. I was wondering if anyone else had anymore luck finding the regulated NOx emissions limit (in ppm or otherwise) for power generating plants in California? Any help would be appreciated. Cheers 150.49.180.199 (talk) 02:50, 15 February 2011 (UTC)

Decompressing the spine

Is there a reason that some people who have had a back injury feel their spine get compressed throughout the day and feel better when they decompress it with an inversion table and people who have not had a back injury do not feel like their spine is getting compressed? —Preceding unsigned comment added by 76.169.33.234 (talk) 06:51, 15 February 2011 (UTC)

The spine gets compressed during the day by standing, in either case, but it may be more of a problem for those with a spinal injury, where the compression may result in pain. StuRat (talk) 08:01, 15 February 2011 (UTC)

That's right: one doesn't need to have had a back injury to feel the results of the compression (OR). Dbfirs 08:15, 15 February 2011 (UTC)

If you don't have a back injury, you probably don't care about it, but the compression reduces your height in about 1 inch along the day. Quest09 (talk) 15:32, 15 February 2011 (UTC)

SPEED OF LIGHT

I am a student of science and want to know that light has a speed of 3*10^8 m/s. Why not greater than it?Mohammad babar (talk) 07:38, 15 February 2011 (UTC)

http://en.wikipedia.org/wiki/Meter#Speed_of_light 157.193.175.207 (talk) 08:13, 15 February 2011 (UTC)

It is what it is because of the way the Second and the Metre are defined. To "increase" the speed of light (which is a universal constant) you would have to either shorten the metre or lengthen the second. Roger (talk) 09:03, 15 February 2011 (UTC)

That (the definition of the meter in terms of the speed of light) is of course a very recent definition, and both the meter and measurements of the speed of light existed before that definition. OP's question makes more sense if turned around: Given the speed of light (and the values of the other constants of nature), why do atoms and people have the sizes that they have? That is a question that can actually be answered. The values of the constants of nature are contingent (at least in the context of our present theories), we do not know of any reason why they are what they are. --Wrongfilter (talk) 09:14, 15 February 2011 (UTC)

This page claims that if the speed of light was higher, nuclear decay would be faster and nuclear reactions in stars would be faster. Therefore the sun might have gone out by now, even if it hadn't the Earth would be much colder (because it gets some of its heat from nuclear decay) and the universe would last a much shorter period of time. Possibly this would be too little time for human beings to evolve - see anthropic principle - but that is just my speculation. --Colapeninsula (talk) 10:13, 15 February 2011 (UTC)

This is a valid argument that the values are such as they are (inferred from the observation that we exist). It is (scientifically) not acceptable as an argument why they are such as they are. --Wrongfilter (talk) 11:26, 15 February 2011 (UTC)

Actually, that is debatable. See anthropic principle. If the constants were significantly different, we wouldn't be here to observe them. Or, in other words, the fact that the universe can sustain life is a precondition for us being able to observe it. --Stephan Schulz (talk) 12:05, 15 February 2011 (UTC)

Which is exactly what I'm saying. From the observation of our existence we can infer what the values are, because if they were different we wouldn't be here. But we cannot claim that our existence is the reason or the cause (in some teleological sense) for their existence. They were not chosen such that we might come into existence. --Wrongfilter (talk) 12:52, 15 February 2011 (UTC)

Is Mohammad actually seeking an explanation of why light has a maximum speed? You may find the article Introduction to special relativity useful, or maybe the Simple English Wikipedia article, which you can read by clicking here. If you tell us a language you prefer, we can show you explanations in that language. 86.164.25.178 (talk) 11:58, 15 February 2011 (UTC)

You can always ask that question though. If the speed of light was 100 mph faster, you could then ask why can't it be faster than that? If the question is, why is the speed of light what it is, I'm not sure we have an explanation of why it goes at that speed. That's just what we observe its speed to be. Obviously, reading about Special Relativity should answer some other questions. We do know that light travels slower through a medium like air or water. Also the speed of light may have been different in the past, maybe someone can expound on that. ScienceApe (talk) 15:09, 15 February 2011 (UTC)

somthing about the creation theory of earth

I know that this is not the place to say my own ideas but I ask an opportunity for saying about only this subject that refers to earth. excuse me . I want to know can i do so? Akbar mohammadzade Iran This is the first time that my own idea is publishing and for the reason of what I think about it is my patent ,may be occure some difficulties in understanding that subject and i have to do so . I am trying to publish my theory .when it happen, I will say what I am thinking about the solar system specially about the earth . According to the data bases collected and the reality observations in astronomy and geology ,it is clear that the earth , mars , venous and mercury which known rock planets created several years after the creation of sun and gas giant planets :Jupiter , Uranus ,Saturn and Neptun. the last bodies of solar system are moons ,Pluto and sates , comas and…. Which has less age than earth . I am trying to upgrade recent astrophysics theories that they are approaching the solution of the paradoxes in theory and practice .--78.38.28.3 (talk) 09:16, 15 February 2011 (UTC)iran feb 2011 mohammadzade

Sooooo what's your question? Someguy1221 (talk) 09:47, 15 February 2011 (UTC)

He asks whether we mind if he tells us his new theory. And I think our advice is probably that he should find an internet forum where people discuss such things, rather than explain it here. Itsmejudith (talk) 16:37, 15 February 2011 (UTC)

Tidal Force

I'm confused why the moon causes a high tide on the opposite side of the earth. The article 'Tidal Force' shows the tidal forces in figures 2 and 4. Could someone explain what "the residual force after the field of the sphere is deducted" means and why it results in a net outward force on the opposite side of the satelite? Thank you80.168.88.74 (talk) 11:47, 15 February 2011 (UTC)

In short, on the side near to the moon, the moon pulls the water harder than the Earth in total (since it's about 6000 km closer than the center of mass of the Earth). On side opposite the moon, the moon pulls the Earth harder than the water (since the water is about 6000 km farther away than the center of mass). Think about it not as "the water is pulled somewhere", think " the Earth is pulled somewhere, the water remains". --Stephan Schulz (talk) 12:02, 15 February 2011 (UTC)

In effect, the Earth is falling towards the Moon, at a speed that it wants to "on average". The parts of Earth nearest to the Moon want to fall faster than this average, so they're tugged away from Earth towards the Moon. The parts furthest from the Moon want to fall more slowly than this average, so they lag begind. 86.181.174.29 (talk) 12:36, 15 February 2011 (UTC)

And this might be clearly not obvious, as we all should know that when objects fall on Earth, it doesn't matter how massive they are - they always fall with the same acceleration. However, this is only because the effective distances from the center (of mass) of the Earth to the objects that we throw in the air are about the same - as the poster above points out, the oceans on each side of the Earth are about 6000km away from the Earth's center of mass, which ends up making a "big" difference.

I put "big" in quotation marks because remember that relative to the 6000km radius of the Earth, a tide of 1 meter is an utterly-miniscule astrophysical effect, unnoticeable if it were watched from another planet. SamuelRiv (talk) 16:34, 15 February 2011 (UTC)

Quark + Gluon plasma - Evidences

Reading article about matter states I could see reference about quark-gluon plasma that was found in CERN in 2000. From other books I have read that it wasn´t possible until now to detect any quark isolated. The evidence that they exist was taken by electrical effects when electrons travel close to the protons and was detected three concentrated points that have strong effect. It was assumed that they were 02 quarks up and 01 down. So is there more evidence in CERN 2000 experiment that confirms that quarks and gluons really exists ? — Preceding unsigned comment added by Futurengineer (talk • contribs) 11:58, 15 February 2011 (UTC)

Schwarzschild Metric

If one sets the angular terms to be zero (by considering a particle travelling on a radial line of the field source) and also sets the spacetime interval to be zero (by considering a photon), the equation for dr/dt does not give c. Why? —Preceding unsigned comment added by 129.67.37.227 (talk) 13:43, 15 February 2011 (UTC)

In general relativity the coordinates are arbitrary and meaningless except as interpreted by the metric. So the t coordinate is not true time, nor is the r coordinate a true distance. They are just numbers that label events in the spacetime, and in order to find distances and times you need to use the metric.

Alternatively, find a local coordinate transformation that makes the metric into the diag(1,1,1,-1) of SR. Then in that frame your photon will move with speed c (tautologically because you selected it to have null worldline). –Henning Makholm (talk) 14:22, 15 February 2011 (UTC)

Discontinuing a vaccine

Isn't it always to risky to stop immunizing your population against a well-known illness? Considering that there are reclusive states which could be, intending or not, a source of a new outbreak. Quest09 (talk) 15:37, 15 February 2011 (UTC)

The risks of the immunization procedure itself must also be taken into consideration and balanced against the risk of an outbreak. Dauto (talk) 15:47, 15 February 2011 (UTC)

In a practical sense, consider the case of smallpox. It is basically eradicated from the wild, however it still exists in laboratories, and could be weaponized by parties with ill intent. If you were a policymaker, you'd have to balance the risk of smallpox being used as an offensive weapon against the risks inherent to the vaccine. Smallpox vaccines have a rate of complications/infections of something like 14-500 per million according to our article. So in a country of 300 million people, that is a fairly large "peacetime" number of people suffering (it would be a justified number compared to that of a smallpox outbreak when it was still "wild"). Whether or not that is a good bet with regards to the threat of smallpox as an agent of war or terrorism is unclear; the US has evidently decided it is not, and does not vaccinate against it generally. Note that in the case of a weaponized threat, you can also say that you might have non-vaccination means of prevention, e.g. a state of deterrence with other nations, or vigorous counter-terrorism strategies. Each of these approaches have their own risk levels and possible numbers of incidental deaths (e.g. you will have a certain number of false positives with your counter-terrorism strategies that may lead to collateral deaths, infringements of liberties, etc.). In all of these situations, people have to make judgments about relative risks and balance accordingly. --Mr.98 (talk) 17:42, 15 February 2011 (UTC)

Yes, see also herd immunity and anti-vaccine conspiracy theories. There has been opposition to vaccines by some, ever since they have been invented. And there have been actual cases of actual problems with vaccines. As a general rule, though, the anti-vaccination crowd is full of nutters, motivation by religion or philosophy, not by the weight of evidence. Friday (talk) 17:03, 15 February 2011 (UTC)

Chemical equation

alkyne (CnH2n-2) + Hydrogen sulfide (H2S) = ???????????????????? Rahul^{Text me} 17:19, 15 February 2011 (UTC)